Maharashtra Board Class 12 Physics Chapter 14 Dual Nature of Radiation and Matter Exercise Solutions

Question. A photocell is used to automatically switch on the street lights in the evening when the sunlight is low in intensity. Thus it has to work with visible light. The material of the cathode of the photocell is
(a) zinc
(b) aluminum
(c) nickel
(d) potassium
Answer: (d) potassium
In simple words: Potassium has the lowest work function, so it can emit electrons even with visible light which has lower energy than UV light.

📝 Teacher's Note: Show students the work function table and explain that visible light photons have less energy than UV, so we need a metal with very low work function. Use the analogy of a low fence being easier to jump over.

🎯 Exam Tip: Remember that for visible light applications, always choose the metal with the lowest work function from the given options - usually potassium or cesium.

Question. Polychromatic (containing many different frequencies) radiation is used in an experiment on the photoelectric effect. The stopping potential
(a) will depend on the average wavelength
(b) will depend on the longest wavelength
(c) will depend on the shortest wavelength
(d) does not depend on the wavelength
Answer: (c) will depend on the shortest wavelength
In simple words: The shortest wavelength has the highest energy and creates the fastest electrons, which determine the stopping potential needed.

📝 Teacher's Note: Emphasize that stopping potential is determined by the most energetic photoelectrons. Use the analogy of the fastest runner in a race setting the winning time.

🎯 Exam Tip: For polychromatic radiation questions, always remember: shortest wavelength → highest frequency → highest energy → determines stopping potential.

Question. An electron, a proton, an α-particle and a hydrogen atom are moving with the same kinetic energy. The associated de Broglie wavelength will be longest for
(a) electron
(b) proton
(c) α-particle
(d) hydrogen atom
Answer: (a) electron
In simple words: Since \( \lambda = \frac{h}{p} \) and \( p = \sqrt{2mKE} \), the electron has the smallest mass, so smallest momentum, so longest wavelength.

📝 Teacher's Note: Derive the relationship \( \lambda = \frac{h}{\sqrt{2mKE}} \) to show wavelength is inversely proportional to square root of mass. Compare masses: electron << proton ≈ H atom < α-particle.

🎯 Exam Tip: For same kinetic energy, lighter particle has longer de Broglie wavelength. Electron is always the lightest among common particles.

Question. If \( N_{Red} \) and \( N_{Blue} \) are the number of photons emitted by the respective sources of equal power and equal dimensions in unit time, then
(a) \( N_{Red} < N_{Blue} \)
(b) \( N_{Red} = N_{Blue} \)
(c) \( N_{Red} > N_{Blue} \)
(d) \( N_{Red} \approx N_{Blue} \)
Answer: (c) \( N_{Red} > N_{Blue} \)
In simple words: Red light has lower energy per photon than blue light, so for the same total power, red sources emit more photons.

📝 Teacher's Note: Use the analogy of coins: if you have the same amount of money to spend, you can buy more 1-rupee coins than 5-rupee coins. Red photons are like cheaper coins.

🎯 Exam Tip: For equal power sources: Power = Number of photons × Energy per photon. Lower energy photons (red) means higher number for same power.

Question. The equation \( E = pc \) is valid
(a) for all sub-atomic particles
(b) is valid for an electron but not for a photon
(c) is valid for a photon but not for an electron
(d) is valid for both an electron and a photon
Answer: (c) is valid for a photon but not for an electron
In simple words: This equation is only for massless particles like photons. Electrons have mass, so they follow \( E = \sqrt{(pc)^2 + (mc^2)^2} \).

📝 Teacher's Note: Distinguish between relativistic energy equations for massless (E = pc) and massive particles. Show that for electrons, rest mass energy cannot be ignored.

🎯 Exam Tip: \( E = pc \) is exclusively for photons and other massless particles. For electrons, always use the full relativistic energy formula.

Question. What is photoelectric effect?
Answer: The phenomenon of emission of electrons from a metal surface when electromagnetic radiation of appropriate frequency is incident on it is known as photoelectric effect.
In simple words: When light of high enough frequency hits a metal surface, it can knock out electrons from the metal.

📝 Teacher's Note: Demonstrate with a simple analogy of throwing balls at a wall to knock off stuck objects. Emphasize the threshold frequency concept - not all light can cause emission.

🎯 Exam Tip: Always mention "appropriate frequency" or "frequency above threshold" in your definition to get full marks.

Question. Can microwaves be used in the experiment on photoelectric effect?
Answer: No
In simple words: Microwaves have very low frequency and energy, far below the threshold needed to knock electrons out of metals.

📝 Teacher's Note: Compare the electromagnetic spectrum - microwaves are much lower in frequency than visible light, which itself is often insufficient for photoelectric effect.

🎯 Exam Tip: For photoelectric effect, you typically need UV light or higher frequency radiation - visible light works only with very reactive metals.

Question. Is it always possible to see photoelectric effect with red light?
Answer: No
In simple words: Red light has low energy and most metals have work functions higher than red light's photon energy.

📝 Teacher's Note: Show students that red light can only cause photoelectric effect in very reactive metals like cesium or potassium, not in common metals like zinc or copper.

🎯 Exam Tip: Red light has energy around 1.8 eV - compare this with work function values in the data table to determine if photoelectric effect is possible.

Question. Using the values of work function given in Table 14.1, tell which metal will require the highest frequency of incident radiation to generate photocurrent.

Answer: Gold.
[ Note : \( W_0 = h\nu_0 \), where h is Planck's constant. The larger the work function (\( W_0 \)), the higher is the threshold frequency (\( \nu_0 \)). ]
In simple words: Gold has the highest work function (5.1 eV), so it needs the highest frequency light to start emitting electrons.

📝 Teacher's Note: Relate work function to the binding energy of electrons - stronger binding requires more energy to overcome. Gold atoms hold their electrons most tightly among the given metals.

🎯 Exam Tip: Always look for the highest work function value in the table - that metal will need the highest threshold frequency according to \( h\nu_0 = W_0 \).

Question. What do you understand by the term wave-particle duality? Where does it apply?
Answer: Depending upon experimental conditions or structure of matter, electromagnetic radiation and material particles exhibit wave nature or particle nature. This is known as wave-particle duality.

It applies to all phenomena. The wave nature and particle nature are linked by the de Broglie relation \( \lambda = h/p \), where \( \lambda \) is the wavelength of matter waves, also called de Broglie waves / Schrodinger waves, p is the magnitude of the momentum of a particle or quantum of radiation and h is the universal constant called Planck's constant.

[Note : It is the smallness of h (= 6.63 × 10⁻³⁴ J∙s) that is very significant in wave-particle duality.]
In simple words: All particles and radiation can behave like waves or particles depending on the experiment, connected by the de Broglie formula.

📝 Teacher's Note: Use examples like electron diffraction (wave nature) and photoelectric effect (particle nature) to illustrate how the same entity shows different behaviors in different experiments.

🎯 Exam Tip: Always mention that duality applies to ALL matter and radiation, and include the de Broglie relation \( \lambda = h/p \) in your answer.

Question. A photocell is used to automatically switch on the street lights in the evening when the sunlight is low in intensity. Thus it has to work with visible light. The material of the cathode of the photocell is
(a) zinc
(b) aluminum
(c) nickel
(d) potassium
Answer: (d) potassium
In simple words: Potassium has the lowest work function among these metals, so it can easily emit electrons even with visible light of lower energy.

📝 Teacher's Note: Demonstrate this concept by showing students the work function table - metals with lower work function need less energy to emit electrons. Potassium works with visible light while zinc needs UV light.

🎯 Exam Tip: Remember: lower work function = works with visible light. Potassium and cesium are commonly used in photocells for this reason.

Question. Polychromatic (containing many different frequencies) radiation is used in an experiment on the photoelectric effect. The stopping potential
(a) will depend on the average wavelength
(b) will depend on the longest wavelength
(c) will depend on the shortest wavelength
(d) does not depend on the wavelength
Answer: (c) will depend on the shortest wavelength
In simple words: The most energetic electrons come from the highest frequency (shortest wavelength) light, so the stopping potential depends on the shortest wavelength present.

📝 Teacher's Note: Explain that stopping potential stops the fastest electrons - these come from the highest energy photons (shortest wavelength). Other wavelengths produce slower electrons that are stopped at lower potentials.

🎯 Exam Tip: Key insight: stopping potential is determined by the maximum kinetic energy, which comes from the highest frequency component in polychromatic light.

Question. An electron, a proton, an α-particle and a hydrogen atom are moving with the same kinetic energy. The associated de Broglie wavelength will be longest for
(a) electron
(b) proton
(c) α-particle
(d) hydrogen atom
Answer: (a) electron
In simple words: Since \( \lambda = \frac{h}{p} \) and \( p = \sqrt{2mKE} \), for same kinetic energy, the lightest particle (electron) has the longest wavelength.

📝 Teacher's Note: Use the analogy of throwing balls of different masses with same energy - lighter ball moves faster and has different wave properties. Show the mathematical relationship clearly.

🎯 Exam Tip: Remember: same KE → lighter mass → smaller momentum → longer de Broglie wavelength. Electron is lightest, so longest wavelength.

Question. If \( N_{Red} \) and \( N_{Blue} \) are the number of photons emitted by the respective sources of equal power and equal dimensions in unit time, then
(a) \( N_{Red} < N_{Blue} \)
(b) \( N_{Red} = N_{Blue} \)
(c) \( N_{Red} > N_{Blue} \)
(d) \( N_{Red} \approx N_{Blue} \)
Answer: (c) \( N_{Red} > N_{Blue} \)
In simple words: Red light has lower energy per photon than blue light, so for equal power, more red photons are needed than blue photons.

📝 Teacher's Note: Use the analogy of buying candy - if red candy costs less than blue candy, you can buy more red pieces with the same money. Here, energy is like cost and photons are like candy pieces.

🎯 Exam Tip: Power = Number of photons × Energy per photon. For equal power: lower energy photons → more photons needed. Red < Blue in energy.

Question. The equation \( E = pc \) is valid
(a) for all sub-atomic particles
(b) is valid for an electron but not for a photon
(c) is valid for a photon but not for an electron
(d) is valid for both an electron and a photon
Answer: (c) is valid for a photon but not for an electron
In simple words: This equation works only for massless particles like photons. Electrons have mass, so they need a different energy-momentum relationship.

📝 Teacher's Note: Emphasize that photons are massless and always travel at speed c, while electrons have mass and can travel at various speeds less than c. Different physics applies.

🎯 Exam Tip: \( E = pc \) is for massless particles only. For massive particles like electrons, use \( E^2 = (pc)^2 + (mc^2)^2 \).

Question. What is photoelectric effect?
Answer: The phenomenon of emission of electrons from a metal surface when electromagnetic radiation of appropriate frequency is incident on it is known as photoelectric effect.
In simple words: When light of high enough frequency hits a metal surface, electrons are knocked out of the metal - like knocking balls off a table with a stick.

📝 Teacher's Note: Demonstrate with a zinc plate and UV lamp if available. Emphasize that frequency, not intensity, determines whether electrons are emitted. This was revolutionary in physics.

🎯 Exam Tip: Key points for full marks: mention "appropriate frequency", "emission of electrons", and "from metal surface". Avoid saying "any light" - frequency threshold is crucial.

Question. Can microwaves be used in the experiment on photoelectric effect?
Answer: No
In simple words: Microwaves have very low frequency and energy, far below what's needed to knock electrons out of any metal.

📝 Teacher's Note: Compare frequencies: microwaves (~GHz) vs visible light (~THz). Show students the electromagnetic spectrum and work function values to illustrate the huge energy gap.

🎯 Exam Tip: Simple "No" is sufficient, but if explaining, mention that microwave frequency is much lower than threshold frequency for any known metal.

Question. Is it always possible to see photoelectric effect with red light?
Answer: No
In simple words: Red light has relatively low energy, so it can only cause photoelectric effect in metals with very low work function like potassium or cesium.

📝 Teacher's Note: Reference the work function table - show that red light energy (~1.8 eV) is below the work function of most metals except alkali metals like potassium.

🎯 Exam Tip: "No" is correct. If elaborating, mention that it depends on the metal's work function - most metals require higher energy than red light provides.

Question. Using the values of work function given in Table 14.1, tell which metal will require the highest frequency of incident radiation to generate photocurrent.
Answer: Gold.
[ Note : \( W_0 = hv_0 \), where h is Planck's constant. The larger the work function (\( W_0 \)), the higher is the threshold frequency (\( v_0 \)). ]
In simple words: Gold has the highest work function (5.1 eV), so it needs the highest frequency light to knock out electrons - like needing a harder hit to knock a ball off a high shelf.

📝 Teacher's Note: Go through the table systematically with students. Show that work function directly relates to threshold frequency through \( W_0 = hf_0 \). Gold needs UV light while potassium works with visible light.

🎯 Exam Tip: Always check the given table for work functions. Highest work function = highest threshold frequency. Show the relationship \( f_0 = W_0/h \) if asked for reasoning.

Question. What do you understand by the term wave-particle duality? Where does it apply?
Answer: Depending upon experimental conditions or structure of matter, electromagnetic radiation and material particles exhibit wave nature or particle nature. This is known as wave-particle duality.
It applies to all phenomena. The wave nature and particle nature are linked by the de Broglie relation \( \lambda = h/p \), where \( \lambda \) is the wavelength of matter waves, also called de Broglie waves / Schrodinger waves, p is the magnitude of the momentum of a particle or quantum of radiation and h is the universal constant called Planck's constant.
[Note : It is the smallness of h (= \( 6.63 \times 10^{-34} \) J∙s) that is very significant in wave-particle duality.]
In simple words: Everything can behave like both a wave and a particle depending on how we observe it - like light being both a wave (showing interference) and particles (photons hitting a detector).

📝 Teacher's Note: Use examples students can relate to: light showing interference (wave) but also photoelectric effect (particle). Emphasize that it's not either/or but both aspects exist simultaneously.

🎯 Exam Tip: Mention both electromagnetic radiation AND matter particles show duality. Include the de Broglie relation and state that it applies to all phenomena for full marks.

Question 3. Explain the inverse linear dependence of stopping potential on the incident wavelength in a photoelectric effect experiment.
Answer: We have \( V_0e = \frac{hc}{\lambda} - \Phi \), where \( V_0 \) is the stopping potential, e is the magnitude of the charge on the electron, h is Planck's constant, c is the speed of light in free space, \( \lambda \) is the wavelength of the electromagnetic radiation incident on a metal surface and \( \Phi \) is the work function for the metal, h, c and e are constants. \( \Phi \) is constant for a particular metal.
Hence, it follows that as \( \frac{1}{\lambda} \) increases, \( V_0 \) increases.
The plot of \( V_0 \) verses \( \frac{1}{\lambda} \) is linear. This is because the energy associated with a quantum of radiation (photon) is directly proportional to the frequency of radiation and hence inversely proportional to the wavelength of radiation.
In simple words: Shorter wavelength light has more energy per photon, so it creates faster electrons that need higher stopping voltage - like throwing balls harder makes them go faster.

📝 Teacher's Note: Draw the graph of stopping potential vs 1/λ to show the linear relationship. Emphasize that this relationship was crucial evidence for the quantum nature of light.

🎯 Exam Tip: Write the photoelectric equation in terms of wavelength: \( eV_0 = \frac{hc}{\lambda} - \Phi \). Show that \( V_0 \propto \frac{1}{\lambda} \) for full marks.

Question. (i) A photocell is used to automatically switch on the street lights in the evening when the sunlight is low in intensity. Thus it has to work with visible light. The material of the cathode of the photocell is
(a) zinc
(b) aluminum
(c) nickel
(d) potassium
Answer: (d) potassium
In simple words: Potassium has the lowest work function, so it can emit electrons even with low-energy visible light — perfect for detecting dim sunlight in the evening.

📝 Teacher's Note: Show students the work function table and explain why alkali metals like potassium are ideal for photocells. Demonstrate with examples of different metals and their threshold frequencies.

🎯 Exam Tip: Remember "Potassium for photocells" — its low work function (2.3 eV) makes it sensitive to visible light, unlike metals with higher work functions.

Question. (ii) Polychromatic (containing many different frequencies) radiation is used in an experiment on the photoelectric effect. The stopping potential
(a) will depend on the average wavelength
(b) will depend on the longest wavelength
(c) will depend on the shortest wavelength
(d) does not depend on the wavelength
Answer: (c) will depend on the shortest wavelength
In simple words: The stopping potential depends on the most energetic electrons, which come from the shortest wavelength (highest frequency) light in the mixture.

📝 Teacher's Note: Use the analogy of a race — the stopping potential is determined by the "fastest runner" (highest energy photons), not the average speed of all runners.

🎯 Exam Tip: Write "shortest wavelength = highest frequency = most energetic photons = determines stopping potential" to show clear logical connection.

Question. (iii) An electron, a proton, an α-particle and a hydrogen atom are moving with the same kinetic energy. The associated de Broglie wavelength will be longest for
(a) electron
(b) proton
(c) α-particle
(d) hydrogen atom
Answer: (a) electron
In simple words: Since \( \lambda = \frac{h}{\sqrt{2mKE}} \), the lighter the particle, the longer its wavelength — electrons are the lightest among the given options.

📝 Teacher's Note: Emphasize the inverse relationship between mass and wavelength when kinetic energy is constant. Use mass comparisons: electron < hydrogen atom < proton < α-particle.

🎯 Exam Tip: For same KE, write \( \lambda \propto \frac{1}{\sqrt{m}} \) — lightest particle gets longest wavelength. Always check particle masses first.

Question. (iv) If \( N_{Red} \) and \( N_{Blue} \) are the number of photons emitted by the respective sources of equal power and equal dimensions in unit time, then
(a) \( N_{Red} < N_{Blue} \)
(b) \( N_{Red} = N_{Blue} \)
(c) \( N_{Red} > N_{Blue} \)
(d) \( N_{Red} \approx N_{Blue} \)
Answer: (c) \( N_{Red} > N_{Blue} \)
In simple words: Red light has lower energy per photon than blue light, so for the same total power, you need more red photons than blue photons.

📝 Teacher's Note: Use the money analogy — if you have the same amount to spend, you can buy more low-value coins than high-value coins. Here, power is like total money, photon energy like coin value.

🎯 Exam Tip: Write \( P = NhV \), so \( N \propto \frac{1}{V} \). Lower frequency (red) means more photons for same power.

Question. (v) The equation E = pc is valid
(a) for all sub-atomic particles
(b) is valid for an electron but not for a photon
(c) is valid for a photon but not for an electron
(d) is valid for both an electron and a photon
Answer: (c) is valid for a photon but not for an electron
In simple words: This equation works only for massless particles like photons. Electrons have mass, so they follow \( E^2 = (pc)^2 + (mc^2)^2 \).

📝 Teacher's Note: Explain the difference between relativistic energy equations for massless vs massive particles. Show how E = pc is the limiting case when rest mass is zero.

🎯 Exam Tip: Remember "E = pc for photons only" — if mass ≠ 0, use the full relativistic energy equation \( E^2 = (pc)^2 + (mc^2)^2 \).

Question. What is photoelectric effect?
Answer: The phenomenon of emission of electrons from a metal surface when electromagnetic radiation of appropriate frequency is incident on it is known as photoelectric effect.
In simple words: When light hits a metal surface, it can knock out electrons from the metal, just like a ball hitting marbles off a table.

📝 Teacher's Note: Start with everyday examples like automatic door sensors or solar panels to make the concept relatable before introducing the technical definition.

🎯 Exam Tip: Include keywords "emission of electrons", "metal surface", "electromagnetic radiation", and "appropriate frequency" in your definition.

Question. Can microwaves be used in the experiment on photoelectric effect?
Answer: No
In simple words: Microwaves have very low frequency and energy, not enough to overcome the work function of most metals and eject electrons.

📝 Teacher's Note: Compare microwave frequency (GHz) with visible light frequency (10¹⁴ Hz) to show the huge energy difference. Relate to work function values in eV.

🎯 Exam Tip: State "No, frequency too low" and briefly explain that microwave photon energy is much less than typical work functions (few eV).

Question. Is it always possible to see photoelectric effect with red light?
Answer: No
In simple words: Red light has low energy, so it can only eject electrons from metals with very low work functions like alkali metals, not from most common metals.

📝 Teacher's Note: Show students the work function table and red light energy (~1.8 eV) to demonstrate which metals would work. Emphasize the threshold concept.

🎯 Exam Tip: Explain briefly that red light energy (~1.8 eV) is less than work function of most metals except alkali metals.

Question. Using the values of work function given in Table 14.1, tell which metal will require the highest frequency of incident radiation to generate photocurrent.
Answer: Gold.
[ Note : \( W_0 = h\nu_0 \), where h is Planck's constant. The larger the work function (\( W_0 \)), the higher is the threshold frequency (\( \nu_0 \)). ]
In simple words: Gold has the highest work function (5.1 eV), so it needs the highest energy (highest frequency) light to knock out electrons.

📝 Teacher's Note: Have students scan the work function table to identify the highest value. Connect work function directly to threshold frequency using \( W_0 = h\nu_0 \).

🎯 Exam Tip: Always refer to the data table given. Write "Gold has highest work function (5.1 eV), so highest threshold frequency needed."

Question. What do you understand by the term wave-particle duality? Where does it apply?
Answer: Depending upon experimental conditions or structure of matter, electromagnetic radiation and material particles exhibit wave nature or particle nature. This is known as wave-particle duality.

It applies to all phenomena. The wave nature and particle nature are liked by the de Broglie relation λ = h/p, where λ is the wavelength of matter waves, also called de Broglie waves / Schrodinger waves, p is the magnitude of the momentum of a particle or quantum of radiation and h is the universal constant called Planck's constant.

[Note : It is the smallness of h (= 6.63 × 10⁻³⁴ J∙s) that is very significant in wave-particle duality.]
In simple words: Light and matter can act like waves or particles depending on how you observe them — like a coin that shows different faces when viewed from different angles.

📝 Teacher's Note: Use examples like electron diffraction (wave nature) and photoelectric effect (particle nature) to illustrate duality. Emphasize that it's about the experimental setup, not the object itself changing.

🎯 Exam Tip: Define duality, give de Broglie relation λ = h/p, mention it applies universally, and note the significance of Planck's constant being very small.

Question 3. Explain the inverse linear dependence of stopping potential on the incident wavelength in a photoelectric effect experiment.
Answer: We have \( V_0e = \frac{hc}{\lambda} - \Phi \), where \( V_0 \) is the stopping potential, e is the magnitude of the charge on the electron, h is Planck's constant, c is the speed of light in free space, λ is the wavelength of the electromagnetic radiation incident on a metal surface and Φ is the work function for the metal, h, c and e are constants. Φ is constant for a particular metal.

Hence, it follows that as \( \frac{1}{\lambda} \) increases, \( V_0 \) increases.

The plot of \( V_0 \) verses \( \frac{1}{\lambda} \) is linear. This is because the energy associated with a quantum of radiation (photon) is directly proportional to the frequency of radiation and hence inversely proportional to the wavelength of radiation.
In simple words: Shorter wavelength means higher energy photons, which create faster electrons that need a higher stopping potential to stop them — it's a straight-line relationship.

📝 Teacher's Note: Graph \( V_0 \) vs \( \frac{1}{\lambda} \) on the board to show linearity. Connect the slope to fundamental constants and emphasize the inverse relationship between wavelength and photon energy.

🎯 Exam Tip: Start with the photoelectric equation, rearrange to show \( V_0 \propto \frac{1}{\lambda} \), and explain the physical meaning of shorter wavelength → higher energy → higher stopping potential.

Question 1. Choose the correct answer.

(i) A photocell is used to automatically switch on the street lights in the evening when the sunlight is low in intensity. Thus it has to work with visible light. The material of the cathode of the photocell is
(a) zinc
(b) aluminum
(c) nickel
(d) potassium
Answer: (d) potassium
In simple words: Potassium has a low work function, meaning it can easily release electrons even when hit by visible light, making it perfect for street light sensors.

📝 Teacher's Note: Show students the work function table to demonstrate why potassium (2.3 eV) is better than other metals for visible light detection. Relate this to why street lights work even in dim light conditions.

🎯 Exam Tip: Remember that for visible light applications, choose metals with low work functions like alkali metals (potassium, sodium, cesium).

(ii) Polychromatic (containing many different frequencies) radiation is used in an experiment on the photoelectric effect. The stopping potential
(a) will depend on the average wavelength
(b) will depend on the longest wavelength
(c) will depend on the shortest wavelength
(d) does not depend on the wavelength
Answer: (c) will depend on the shortest wavelength
In simple words: The most energetic photons (shortest wavelength) produce the fastest electrons, so the stopping potential is determined by these fastest electrons.

📝 Teacher's Note: Use the analogy of a race where only the fastest runner determines the finish line barrier height. The shortest wavelength photons are like the fastest runners.

🎯 Exam Tip: For polychromatic light, always look for the highest frequency (shortest wavelength) component to determine maximum kinetic energy and stopping potential.

(iii) An electron, a proton, an α-particle and a hydrogen atom are moving with the same kinetic energy. The associated de Broglie wavelength will be longest for
(a) electron
(b) proton
(c) α-particle
(d) hydrogen atom
Answer: (a) electron
In simple words: Since \( \lambda = \frac{h}{\sqrt{2mKE}} \), the lightest particle (electron) will have the longest wavelength when all have the same energy.

📝 Teacher's Note: Emphasize that wavelength is inversely proportional to the square root of mass. Use the formula to show numerical comparison between electron and proton masses.

🎯 Exam Tip: For same kinetic energy, lighter particles have longer de Broglie wavelengths. Electron mass is smallest among given options.

(iv) If \( N_{Red} \) and \( N_{Blue} \) are the number of photons emitted by the respective sources of equal power and equal dimensions in unit time, then
(a) \( N_{Red} < N_{Blue} \)
(b) \( N_{Red} = N_{Blue} \)
(c) \( N_{Red} > N_{Blue} \)
(d) \( N_{Red} \approx N_{Blue} \)
Answer: (c) \( N_{Red} > N_{Blue} \)
In simple words: Red photons have less energy than blue photons, so you need more red photons to make the same total power.

📝 Teacher's Note: Use the power equation P = NhE to show that N = P/(hf). Since red light has lower frequency, more photons are needed for same power.

🎯 Exam Tip: For equal power sources: lower frequency light requires more photons. Remember P = Nhf, so N ∝ 1/f.

(v) The equation E = pc is valid
(a) for all sub-atomic particles
(b) is valid for an electron but not for a photon
(c) is valid for a photon but not for an electron
(d) is valid for both an electron and a photon
Answer: (c) is valid for a photon but not for an electron
In simple words: This equation only works for massless particles like photons. For electrons (which have mass), you need the relativistic energy equation.

📝 Teacher's Note: Explain that E = pc is for massless particles, while massive particles use E² = (pc)² + (mc²)². Show how photon mass = 0 simplifies the equation.

🎯 Exam Tip: E = pc is exclusively for photons (massless). For massive particles like electrons, use E = (1/2)mv² for non-relativistic or full relativistic formula.

Question 2. Answer in brief.

(i) What is photoelectric effect?
Answer: The phenomenon of emission of electrons from a metal surface when electromagnetic radiation of appropriate frequency is incident on it is known as photoelectric effect.
In simple words: When light of the right frequency hits a metal surface, it knocks out electrons from the metal, like a key unlocking a door.

📝 Teacher's Note: Demonstrate with a simple analogy of billiard balls hitting other balls. Emphasize that frequency, not intensity, determines if electrons are emitted.

🎯 Exam Tip: Always mention "appropriate frequency" in the definition - this is crucial for full marks. Don't just say "light hits metal."

(ii) Can microwaves be used in the experiment on photoelectric effect?
Answer: No
In simple words: Microwaves have very low frequency and energy, not enough to knock electrons out of most metals used in photoelectric experiments.

📝 Teacher's Note: Compare microwave frequency (around 10¹⁰ Hz) with visible light (around 10¹⁵ Hz) to show the huge energy difference using E = hf.

🎯 Exam Tip: Remember the electromagnetic spectrum order. Microwaves have much lower frequency than visible light, hence insufficient energy for photoelectric effect.

(iii) Is it always possible to see photoelectric effect with red light?
Answer: No
In simple words: Red light has low energy, so it can only cause photoelectric effect in metals with very low work functions like alkali metals.

📝 Teacher's Note: Show students why red light works with cesium but not with zinc by comparing photon energy (1.8 eV) with work functions from the table.

🎯 Exam Tip: Consider both photon energy and work function of the metal. Red light energy ≈ 1.8 eV, which is less than work function of many metals.

(iv) Using the values of work function given in Table 14.1, tell which metal will require the highest frequency of incident radiation to generate photocurrent.
Answer: Gold.
[ Note : \( W_0 = hv_0 \), where h is Planck's constant. The larger the work function (\( W_0 \)), the higher is the threshold frequency (\( v_0 \)). ]
In simple words: Gold has the highest work function (5.1 eV), so you need the most energetic photons (highest frequency) to remove electrons from it.

📝 Teacher's Note: Use the work function table to show the direct relationship between work function and threshold frequency. Gold requires UV light while potassium works with visible light.

🎯 Exam Tip: Higher work function = higher threshold frequency needed. Look for the metal with maximum work function value in the given table.

(v) What do you understand by the term wave-particle duality? Where does it apply?
Answer: Depending upon experimental conditions or structure of matter, electromagnetic radiation and material particles exhibit wave nature or particle nature. This is known as wave-particle duality.

It applies to all phenomena. The wave nature and particle nature are linked by the de Broglie relation \( \lambda = h/p \), where \( \lambda \) is the wavelength of matter waves, also called de Broglie waves / Schrodinger waves, p is the magnitude of the momentum of a particle or quantum of radiation and h is the universal constant called Planck's constant.

[Note : It is the smallness of h (= \( 6.63 \times 10^{-34} \) J∙s) that is very significant in wave-particle duality.]
In simple words: Everything in nature can behave like both a wave and a particle depending on how we observe it, like light being both a wave and made of particles called photons.

📝 Teacher's Note: Use everyday analogies like water waves vs water droplets. Emphasize that this isn't "sometimes wave, sometimes particle" but rather complementary descriptions of the same reality.

🎯 Exam Tip: Always mention the de Broglie relation λ = h/p and state that it applies to ALL matter and radiation, not just specific particles.

Question 3. Explain the inverse linear dependence of stopping potential on the incident wavelength in a photoelectric effect experiment.
Answer: We have \( V_0e = \frac{hc}{\lambda} - \Phi \), where \( V_0 \) is the stopping potential, e is the magnitude of the charge on the electron, h is Planck's constant, c is the speed of light in free space, \( \lambda \) is the wavelength of the electromagnetic radiation incident on a metal surface and \( \Phi \) is the work function for the metal, h, c and e are constants. \( \Phi \) is constant for a particular metal.

Hence, it follows that as \( \frac{1}{\lambda} \) increases, \( V_0 \) increases.

The plot of \( V_0 \) verses \( \frac{1}{\lambda} \) is linear. This is because the energy associated with a quantum of radiation (photon) is directly proportional to the frequency of radiation and hence inversely proportional to the wavelength of radiation.
In simple words: Shorter wavelength light has more energy, which creates faster electrons, so you need higher voltage to stop them - hence stopping potential increases as wavelength decreases.

📝 Teacher's Note: Draw the graph of V₀ vs 1/λ to show the linear relationship. Emphasize that slope = hc/e gives a way to measure Planck's constant experimentally.

🎯 Exam Tip: Remember the key equation V₀e = hc/λ - Φ. Show that V₀ ∝ 1/λ clearly and mention the linear graph relationship for full marks.

Question 4. It is observed in an experiment on photoelectric effect that an increase in the intensity of the incident radiation does not change the maximum kinetic energy of the electrons. Where does the extra energy of the incident radiation go? Is it lost? State your answer with explanatory reasoning.
Answer: When electromagnetic radiation with frequency greater than the threshold frequency is incident on a metal surface, there is emission of electrons. It is observed that not every incident photon is effective in liberating an electron. In fact, the number of electrons emitted per second is far less than the number of photons incident per second. The photons that are not effective in liberation of electrons are reflected (or scattered) or absorbed resulting in rise in the temperature of the metal surface. The maximum kinetic energy of a photoelectron depends on the frequency of the incident radiation and the threshold frequency for the metal. It has nothing to do with the intensity of the incident radiation. The increase in intensity results in increase in the number of electrons emitted per second.
In simple words: The extra energy isn't lost - it either bounces off the metal surface or heats up the metal. More intense light just means more electrons are knocked out, not faster electrons.

📝 Teacher's Note: Use the analogy of many basketballs vs one bowling ball hitting pins. More basketballs (higher intensity) knock out more pins but don't make individual pins fly faster.

🎯 Exam Tip: Clearly state that energy is conserved - extra energy goes into heating the metal surface and increasing photocurrent, not into making electrons faster.

Question 5. Explain what do you understand by the de Broglie wavelength of an electron. Will an electron at rest have an associated de Broglie wavelength? Justify your answer.
Answer: Under certain conditions an electron exhibits wave nature. Waves associated with a moving electron are called matter waves or de Broglie waves or Schrodinger waves. The de Broglie wavelength of these matter waves is given by \( \lambda = h/p \), where h is Planck's constant and p is the magnitude of the momentum of the electron.

If an electron is at rest, its momentum would be zero, and hence the corresponding de Broglie wavelength would be infinite indicating absence of a matter wave. However, according to quantum mechanics/wave mechanics, this is not possible.
In simple words: Moving electrons have a wave-like property with a specific wavelength. An electron at rest would have infinite wavelength, which doesn't make physical sense.

📝 Teacher's Note: Explain that electrons are never truly at rest due to quantum uncertainty principle. Use the uncertainty principle ΔxΔp ≥ h/4π to show why stationary electrons are impossible.

🎯 Exam Tip: Always mention that λ = h/p and for rest electron p = 0, so λ = ∞, which is physically meaningless. Connect this to quantum mechanics principles.

Question 6. State the importance of Davisson and Germer experiment.
Answer: The Davisson and Germer experiment directly indicated the wave nature of material particles and quantitatively verified the de Broglie hypothesis for the existence of matter waves.

[Note : The aim of the experiment was not to verify wave like properties of electrons. The realisation came only later, an example of serendipity.]

[Note : Like X-rays, electrons exhibit wave nature under suitable conditions. When the wavelength of matter waves associated with moving electrons is comparable to the inter-atomic spacing in a crystal, electrons show diffraction effects. In 1927, Sir George Thomson (1892 – 1975), British physicist, with his student Alex Reid, observed electron diffraction with a metal foil. It is found that neutrons, atoms, molecules, α-particles, etc. show wave nature under suitable conditions.]
In simple words: This experiment proved that electrons behave like waves by showing they can create diffraction patterns, just like light waves do.

📝 Teacher's Note: Emphasize the accidental nature of this discovery - they were studying electron scattering but discovered diffraction. This shows how science sometimes progresses through unexpected observations.

🎯 Exam Tip: Key points: (1) First direct evidence of electron wave nature, (2) Verified de Broglie hypothesis quantitatively, (3) Showed diffraction patterns similar to X-rays.

Question 7. What will be the energy of each photon in monochromatic light of frequency \( 5 \times 10^{14} \) Hz?
Answer: Data: \( v = 5 \times 10^{14} \) Hz, \( h = 6.63 \times 10^{-34} \) Js, 1eV = \( 1.6 \times 10^{-19} \) J

The energy of each photon,
E = hv = \( (6.63 \times 10^{-34} \) J.s)(\( 5 \times 10^{14} \) Hz)
= \( 3.315 \times 10^{-19} \) J
= \( \frac{3.315 \times 10^{-19} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} \) = 2.072 eV
In simple words: Each photon of this light carries about 2 eV of energy, which is enough to cause photoelectric effect in metals with work function less than 2 eV.

📝 Teacher's Note: Show students how to convert between Joules and electron volts. This frequency corresponds to green light in the visible spectrum.

🎯 Exam Tip: Always use E = hf and remember to convert final answer to eV using 1 eV = 1.6 × 10⁻¹⁹ J. Show all steps clearly.

Question 8. Observations from an experiment on photoelectric effect for the stopping potential by varying the incident frequency were plotted. The slope of the linear curve was found to be approximately \( 4.1 \times 10^{-15} \) V s. Given that the charge of an electron is \( 1.6 \times 10^{-19} \) C, find the value of the Planck's constant h.
Answer: Data : Slope = \( 4.1 \times 10^{-15} \) V∙s, e = \( 1.6 \times 10^{-19} \) C

\( V_0e = hv - hv_0 \)

\( \therefore V_0 = \frac{h}{e}(v - v_0) \)

\( \therefore \) Slope = \( \frac{h}{e} \)

\( \therefore \) Planck's constant,
h = (slope) (e) = \( (4.1 \times 10^{-15} \) V∙s)(\( 1.6 \times 10^{-19} \) C)
= \( 6.56 \times 10^{-34} \) J∙s (as 1 V = \( \frac{1 \text{ J}}{1 \text{ C}} \))
In simple words: By measuring how stopping voltage changes with frequency, we can calculate Planck's constant, which comes out very close to the accepted value.

📝 Teacher's Note: Emphasize this is how Planck's constant was historically measured. The slope of V₀ vs f graph directly gives h/e.

🎯 Exam Tip: Remember that slope of stopping potential vs frequency graph equals h/e. Multiply slope by electronic charge to get Planck's constant.

Question 9. The threshold wavelength of tungsten is \( 2.76 \times 10^{-5} \) cm. (a) Explain why no photoelectrons are emitted when the wavelength is more than \( 2.76 \times 10^{-5} \) cm.(b) What will be the maximum kinetic energy of electrons ejected in each of the following cases
(i) if ultraviolet radiation of wavelength λ = \( 1.80 \times 10^{-5} \) cm and
(ii) radiation of frequency \( 4 \times 10^{15} \) Hz is made incident on the tungsten surface.
Answer: Data: \( \lambda_0 = 2.76 \times 10^{-5} \) cm = \( 2.76 \times 10^{-7} \) m,
λ = \( 1.80 \times 10^{-5} \) cm = \( 1.80 \times 10^{-7} \) m,
v = \( 4 \times 10^{15} \) Hz, h = \( 6.63 \times 10^{-34} \) J∙s, c = \( 3 \times 10^8 \) m/s

(a) For λ > \( \lambda_0 \), v < \( v_0 \) (threshold frequency).
\( \therefore \) hv < \( hv_0 \). Hence, no photoelectrons are emitted.

(b) Maximum kinetic energy of electrons ejected
= hc \( (\frac{1}{\lambda} - \frac{1}{\lambda_0}) \)
= \( (6.63 \times 10^{-34})(3 \times 10^8)(\frac{10^7}{1.8} - \frac{10^7}{2.76}) \) J
= \( (6.63 \times 10^{-19})(0.5555 - 0.3623) \)
= \( (6.63)(0.1932 \times 10^{-19}) \)J = \( 1.281 \times 10^{-19} \) J
= \( \frac{1.281 \times 10^{-19} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} \) = 0.8006 eV

(c) Maximum kinetic energy of electrons ejected
= hv – \( \frac{hc}{\lambda_0} \)
= \( (6.63 \times 10^{-34})(4 \times 10^{15}) \) – \( \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{2.76 \times 10^{-7}} \)
= \( 26.52 \times 10^{-19} \) – \( 7.207 \times 10^{-19} \)
= \( 19.313 \times 10^{-19} \) J
= \( \frac{19.313 \times 10^{-19} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} \) = 12.07eV
In simple words: Longer wavelength than threshold means lower energy than needed to remove electrons. The UV light gives electrons 0.8 eV while the higher frequency radiation gives 12 eV.

📝 Teacher's Note: Use energy diagrams to show threshold concept. Emphasize that part (ii) gives much higher energy because the frequency is much higher than threshold.

🎯 Exam Tip: For threshold wavelength problems, use KE = hc(1/λ - 1/λ₀) for given wavelength and KE = hf - hc/λ₀ for given frequency. Always convert to eV.

Question 10. Photocurrent recorded in the micro ammeter in an experimental set-up of photoelectric effect vanishes when the retarding potential is more than 0.8 V if the wavelength of incident radiation is 4950 Å. If the source of incident radiation is changed, the stopping potential turns out to be 1.2 V. Find the work function of the cathode material and the wavelength of the second source.
Answer: Data: \( V_0 = 0.8 \) V, λ = 4950 Å = \( 4.950 \times 10^{-7} \) m,
\( V_0' = 1.2 \)V, h = \( 6.63 \times 10^{-34} \) J∙s, c = \( 3 \times 10^8 \) m/s.

(i) \( V_0e = hv - \Phi = \frac{hc}{\lambda} - \Phi \)

\( \therefore \) The work function of the cathode material,
\( \phi = \frac{hc}{\lambda} - V_0e \)

= \( \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{4.950 \times 10^{-7}} \) – \( (0.8)(1.6 \times 10^{-19}) \)

= \( 4.018 \times 10^{-19} \) – \( 1.28 \times 10^{-19} \) = \( 2.738 \times 10^{-19} \) J

= \( \frac{2.738 \times 10^{-19} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} \) = 1.711 eV

(ii) \( V_0'e = \frac{hc}{\lambda'} - \phi \)

\( \therefore \frac{hc}{\lambda'} = V_0'e + \phi \)

\( \therefore \) The wavelength of the second source,
\( \lambda' = \frac{hc}{V_0'e + \phi} \)

= \( \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{(1.2)(1.6 \times 10^{-19}) + 2.738 \times 10^{-19}} \)

= \( \frac{19.89 \times 10^{-26}}{4.658 \times 10^{-19}} \) = \( 4.270 \times 10^{-7} \) m = 4270 Å
In simple words: The metal has a work function of 1.7 eV, and the second light source has a shorter wavelength (4270 Å) because it produces faster electrons.

📝 Teacher's Note: Show how work function is like the "escape energy" for electrons. Higher stopping potential means more energetic photons, hence shorter wavelength.

🎯 Exam Tip: Use the photoelectric equation consistently: hf = KE + Φ. Remember that higher stopping potential indicates higher photon energy, hence shorter wavelength.

Question 11. Photons of energy 2.0 eV fall on a photocathode and eject photoelectrons. The most energetic electrons emitted in a direction perpendicular to the field move along a circular path of radius 20 cm. What is the value of the magnetic field B?
Answer: Data: λ = 4500Å = 4.5 × 10⁻⁷ m,
Φ = 2.0eV = 2 × 1.6 × 10⁻¹⁹ J = 3.2 × 10⁻¹⁹ J,
h = 6.63 × 10⁻³⁴ J∙s, c = 3 × 10⁸ m/s,
r = 20 cm = 0.2 m, e= 1.6 × 10⁻¹⁹ C,
m = 9.1 × 10⁻³¹ kg

\( \frac{1}{2}mv_{max}^2 \) (KE_max) = \( \frac{hc}{\lambda} - \phi \)

= \( \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{4.5 \times 10^{-7}} - 3.2 \times 10^{-19} \)

= 4.42 × 10⁻¹⁹ − 3.2 × 10⁻¹⁹
= 1.22 × 10⁻¹⁹ J

\( v_{max} = \sqrt{\frac{2 KE_{max}}{m}} = \sqrt{\frac{2 \times 1.22 \times 10^{-19}}{9.1 \times 10^{-31}}} \)

= \( \sqrt{0.2681 \times 10^{12}} = 5.178 \times 10^5 \) m/s

Now, centripetal force, \( \frac{mv_{max}^2}{r} = \) magnetic force,

Bev_max

\( B = \frac{mv_{max}}{re} = \frac{(9.1 \times 10^{-31})(5.178 \times 10^5)}{(0.2)(1.6 \times 10^{-19})} \)

= 1.472 × 10⁻⁵ T

This is the value of the magnetic field.
In simple words: When electrons are ejected by photons and move in a magnetic field, they follow circular paths whose radius depends on their speed and the magnetic field strength.

📝 Teacher's Note: Emphasize that this problem combines photoelectric effect with magnetic force. Students should understand that the centripetal force equals the magnetic force for circular motion.

🎯 Exam Tip: Always check units carefully and remember that the most energetic electrons have kinetic energy equal to photon energy minus work function.

Question 12. Given the following data for incident wavelength and the stopping potential obtained from an experiment on photoelectric effect, estimate the value of Planck's constant and the work function of the cathode material. What is the threshold frequency and corresponding wavelength? What is the most likely metal used for emitter?

Answer: Data: λ = 2536Å = 2.536 × 10⁻⁷ m,
λ' = 3650Å = 3.650 ×10⁻⁷ m, V₀ = 1.95V, V₀' = 0.5V,
c = 3 × 10⁸ m/s, e = 1.6 × 10⁻¹⁹ C

(i) V₀e = \( \frac{hc}{\lambda} \) − Φ and V₀'e = \( \frac{hc}{\lambda'} \) − Φ

\( (V_0 - V_0')e = hc\left(\frac{1}{\lambda} - \frac{1}{\lambda'}\right) \)

\( (1.95 - 0.5)(1.6 \times 10^{-19}) = h(3 \times 10^8)\left(\frac{10^7}{2.536} - \frac{10^7}{3.650}\right) \)

\( 2.32 \times 10^{-19} = h(3 \times 10^{15})(0.3943 - 0.2740) \)

\( h = \frac{2.32 \times 10^{-34}}{0.3609} = 6.428 \times 10^{-34} \) J∙s

This is the value of Planck's constant.

(ii) Φ = \( \frac{hc}{\lambda} \) − V₀e
= \( \frac{(6.428 \times 10^{-34})(3 \times 10^8)}{2.536 \times 10^{-7}} - (1.95)(1.6 \times 10^{-19}) \)

= 7.604 × 10⁻¹⁹ − 3.12 × 10⁻¹⁹ = 4.484 × 10⁻¹⁹ J

= \( \frac{4.484 \times 10^{-19} J}{1.6 \times 10^{-19} J/eV} = 2.803 \) eV

This is the work function of the cathode material.

(iii) Φ = hv₀
\( v_0 = \frac{\phi}{h} = \frac{4.484 \times 10^{-19} J}{6.428 \times 10^{-34} J \cdot s} = 6.976 \times 10^{14} \) Hz

(iv) v₀ = \( \frac{c}{\lambda_0} \) \( \lambda_0 = \frac{c}{v_0} = \frac{3 \times 10^8}{6.976 \times 10^{14}} = 4.300 \times 10^{-7} \) m = 4300 Å

(v) The most likely metal used for emitter : calcium
In simple words: By using two different wavelengths and measuring their stopping potentials, we can determine Planck's constant and the work function of the metal.

📝 Teacher's Note: This is a classic method to determine h experimentally. Show students how the slope of V₀ vs. frequency graph gives h/e.

🎯 Exam Tip: When solving photoelectric problems with multiple wavelengths, always use the difference method to eliminate the work function initially.

Question 13. Calculate the wavelength associated with an electron, its momentum and speed (a) when it is accelerated through a potential of 54 V
Answer: Data : V = 54 V, m = 9.1 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C, h = 6.63 × 10⁻³⁴ J.s, KE = 150 eV
(a) We assume that the electron is initially at rest.

Ve = \( \frac{1}{2}mv^2 \)

\( v = \sqrt{\frac{2Ve}{m}} = \sqrt{\frac{2(54)(1.6 \times 10^{-19})}{9.1 \times 10^{-31}}} = \sqrt{19 \times 10^{12}} = 4.359 \times 10^6 \) m/s

This is the speed of the electron.
p = mv = (9.1× 10⁻³¹)(4.359 × 10⁶) = 3.967 × 10⁻²⁴ kg∙m/s

This is the momentum of the electron. The wavelength associated with the electron,
λ = \( \frac{h}{p} = \frac{6.63 \times 10^{-34}}{3.967 \times 10^{-24}} = 1.671 \times 10^{-10} \) m = 1.671 Å = 0.1671 nm

(b) when it is moving with kinetic energy of 150 eV.
Answer: As KE ∝ \( \sqrt{V} \), we get
\( \frac{v'}{v} = \sqrt{\frac{150}{54}} = 1.666 \)

\( v' = 1.666v = (1.666)(4.356 \times 10^6) = 7.262 \times 10^6 \) m/s

This is the speed of the electron.
p' = mv' = (9.1 × 10⁻³¹)(7.262 × 10⁶) = 6.608 × 10⁻²⁴ kg∙m/s

This is the momentum of the electron. The wavelength associated with the electron,
λ = \( \frac{h}{p'} = \frac{6.63 \times 10^{-34}}{6.608 \times 10^{-24}} = 1.003 \times 10^{-10} \) m = 1.003 Å = 0.1003 nm
In simple words: Faster electrons have shorter de Broglie wavelengths, which is why higher energy electrons give better resolution in electron microscopes.

📝 Teacher's Note: Emphasize that de Broglie wavelength is inversely proportional to momentum. Higher kinetic energy means higher speed and momentum, thus shorter wavelength.

🎯 Exam Tip: Remember the relationship λ = h/p and that kinetic energy is proportional to the square of velocity for non-relativistic particles.

Question 14. The de Broglie wavelengths associated with an electron and a proton are same. What will be the ratio of (i) their momenta (ii) their kinetic energies?
Answer: Data : λ (electron) = λ (proton)
m (proton) = 1836 m (electron)
(i) λ = \( \frac{h}{p} \) As λ (electron) = λ (proton),
\( \frac{p(\text{electron})}{p(\text{proton})} = 1 \), where p denotes the magnitude of momentum.

(ii) Assuming v ≪c,
KE = \( \frac{1}{2}mv^2 = \frac{1}{2} \frac{m^2v^2}{m} = \frac{p^2}{2m} \)

\( \frac{KE(\text{electron})}{KE(\text{proton})} = \frac{m(\text{proton})}{m(\text{electron})} = 1836 \) as p is the same for the electron and the proton.
In simple words: When particles have the same wavelength, they have equal momentum, but the lighter particle (electron) has much higher kinetic energy.

📝 Teacher's Note: This problem beautifully shows how momentum and kinetic energy are related differently to mass. Use this to discuss why electrons are more useful in electron microscopes.

🎯 Exam Tip: Remember that equal wavelengths mean equal momenta, but kinetic energy depends inversely on mass when momentum is constant.

Question 15. Two particles have the same de Broglie wavelength and one is moving four times as fast as the other. If the slower particle is an α-particle, what are the possibilities for the other particle?
Answer: Data : λ₁ = λ₂, v₁ = 4v₂
λ = \( \frac{h}{p} = \frac{h}{mv} \) ∴ λ₁ = \( \frac{h}{m₁v₁} \), λ₂ = \( \frac{h}{m₂v₂} \)

∴ m₁ \( \frac{v₂}{v₁} = m₂\left(\frac{1}{4}\right) = \frac{m₂}{4} \)

As particle 2 is the α-particle, particle 1 (having the mass \( \frac{1}{4} \) times that of the α-particle) may be a proton or neutron.
In simple words: Since the faster particle has the same wavelength as the slower α-particle, it must be much lighter—about one-fourth the mass of an α-particle.

📝 Teacher's Note: Help students understand that α-particles are helium nuclei (2 protons + 2 neutrons), so a particle with 1/4 its mass would be a single nucleon.

🎯 Exam Tip: When particles have equal wavelengths but different speeds, use the relationship mv₁ = m₂v₂ to find the mass ratio.

Question 16. What is the speed of a proton having de Broglie wavelength of 0.08 Å?
Answer: Data : λ = 0.08 Å = 8 × 10⁻¹² m, h = 6.63 × 10⁻³⁴ J∙s, m = 1.672 × 10⁻²⁷ kg
λ = \( \frac{h}{mv} \) ∴ v = \( \frac{h}{\lambda m} = \frac{6.63 \times 10^{-34}}{(8 \times 10^{-12})(1.672 \times 10^{-27})} \)

∴ v = 4.957 × 10⁴ m/s

This is the speed of the proton.
In simple words: A proton with this very short wavelength moves at about 50 km/s, which is quite fast but still much slower than light.

📝 Teacher's Note: Compare this speed to everyday speeds to help students appreciate the scale. This is about 180,000 km/h!

🎯 Exam Tip: Always check if the calculated speed is reasonable—it should be much less than the speed of light for non-relativistic calculations.

Question 17. In nuclear reactors, neutrons travel with energies of 5 × 10⁻²¹ J. Find their speed and wavelength.
Answer: Data : KE = 5 × 10⁻²¹ J, m = 1.675 × 10⁻²⁷ kg, h = 6.63 × 10⁻³⁴ J∙s
KE = \( \frac{1}{2}mv^2 = 5 \times 10^{-21} \) J

\( v = \sqrt{\frac{2KE}{m}} = \sqrt{\frac{(2)(5 \times 10^{-21})}{1.675 \times 10^{-27}}} = 2.443 \times 10^3 \) m/s

This is the speed of the neutrons. The de Broglie wavelength associated with the neutron,
λ = \( \frac{h}{mv} = \frac{6.63 \times 10^{-34}}{(1.675 \times 10^{-27})(2.443 \times 10^3)} = 1.620 \times 10^{-10} \) m = 1.620 Å
In simple words: These thermal neutrons move at about 2.4 km/s and have wavelengths similar to X-rays, making them useful for studying crystal structures.

📝 Teacher's Note: Explain that these are "thermal" neutrons used in nuclear reactors and neutron diffraction experiments because their wavelengths match atomic spacings.

🎯 Exam Tip: Remember that neutron wavelengths comparable to atomic spacings (around 1-2 Å) are perfect for studying crystal structures through diffraction.

Question 18. Find the ratio of the de Broglie wavelengths of an electron and a proton when both are moving with the (a) same speed, (b) same energy and (c) same momentum? State which of the two will have the longer wavelength in each case?
Answer: Data: m_p = 1836 m_e

(a) The de Broglie wavelength, λ = \( \frac{h}{p} = \frac{h}{mv} \)

\( \frac{\lambda_e}{\lambda_p} = \frac{m_p}{m_e} \cdot \frac{v_p}{v_e} = 1836 \) as v_p = v_e

Thus, λ_e > λ_p

(b) λ = \( \frac{h}{p} = \frac{h}{\sqrt{2mK}} \), where K denotes the kinetic energy \( \left(\frac{1}{2}mv^2\right) \)

\( \frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p K_p}{m_e K_e}} = \sqrt{\frac{m_p}{m_e}} = \sqrt{1836} = 42.85 \) as K_p = K_e.

Thus, λ_e > λ_p.

(c) λ = \( \frac{h}{p} \)

\( \frac{\lambda_e}{\lambda_p} = \frac{p_p}{p_e} = 1 \) as p_p = p_e.
In simple words: Electrons always have longer wavelengths than protons except when they have equal momentum, in which case the wavelengths are equal.

📝 Teacher's Note: This problem shows three different scenarios. Emphasize that the electron, being lighter, generally has longer wavelengths for the same speed or energy.

🎯 Exam Tip: Remember the three cases: same speed (ratio = mass ratio), same energy (ratio = √mass ratio), same momentum (ratio = 1).

Remember This (Textbook Page No. 316)

Question. Is a solar cell a photocell?
Answer: Yes
In simple words: A solar cell converts light energy directly into electrical energy using the photoelectric effect, so it is indeed a type of photocell.

📝 Teacher's Note: Explain that solar cells use the photovoltaic effect, which is based on the photoelectric effect but designed to generate current rather than just emit electrons.

🎯 Exam Tip: Remember that any device that converts light into electrical signals can be considered a photocell.

Remember This (Textbook Page No. 317)

Question. Can you estimate the de Broglie wavelength of the Earth?
Answer: Taking the mass of the Earth as (about) 6 × 10²⁴ kg, and the linear speed of the earth around the Sun as (about) 3 × 10⁴ m/s, we have, the de Brogue wave length of the Earth as
λ = \( \frac{h}{p} = \frac{h}{Mv} = \frac{6.63 \times 10^{-34} J⋅s}{(6 \times 10^{24} kg)(3 \times 10^4 m/s)} = 3.683 \times 10^{-63} \) m (extremely small)
In simple words: Earth's wavelength is unimaginably small—much smaller than any particle we know, which is why we never observe wave properties of large objects.

📝 Teacher's Note: Use this calculation to show students why wave nature is only observable for very small particles. The wavelength is smaller than anything measurable.

🎯 Exam Tip: This demonstrates why classical mechanics works for large objects—their de Broglie wavelengths are negligibly small.

Question. The expression p = E/c defines the momentum of a photon. Can this expression be used for the momentum of an electron or proton?
Answer: No
In simple words: This formula only works for massless particles like photons. For particles with mass like electrons and protons, we use p = mv.

📝 Teacher's Note: Emphasize that p = E/c applies only to massless particles traveling at light speed. For massive particles, use the classical or relativistic momentum formulas.

🎯 Exam Tip: Never confuse photon momentum (p = E/c) with particle momentum (p = mv). They apply to different types of particles.

Remember This (Textbook Page No. 319)

Diffraction results described above can be produced in the laboratory using an electron diffraction tube as shown in the figure. It has a filament which on heating produces electrons. This filament acts as a cathode. Electrons are accelerated to quite high speeds by creating large potential difference between the cathode and a positive electrode. On its way, the beam of electrons comes across a thin sheet of graphite. The electrons are diffracted by the atomic layers in the graphite and form diffraction rings on the phosphor screen. By changing the voltage between the cathode and anode, the energy, and therefore the speed, of the electrons can be changed. This will change the wavelength of the electrons and a change will be seen in the diffraction pattern. By increasing the voltage, the radius of the diffraction rings will decrease. Try to explain why?

Answer: When the accelerating voltage is increased, the kinetic energy and hence the momentum of the electron increases. This decreases the de Brogue wavelength of the electron. Hence, the radius of the diffraction ring decreases.
In simple words: Higher voltage makes electrons move faster, giving them shorter wavelengths, which creates smaller diffraction patterns.

📝 Teacher's Note: Connect this to the wave nature of electrons. Students should understand that diffraction ring size is inversely related to momentum.

🎯 Exam Tip: Remember that higher energy/momentum means shorter wavelength, which leads to smaller diffraction features.

Remember This (Textbook Page No. 320)

Question. On what scale or under which circumstances are the wave nature of matter apparent?
Answer: When the de Brogue wavelength of a particle such as an electron, atom, or molecule is comparable to the interatomic spacing in a crystal, the wave nature of matter is revealed in diffraction/interference.
In simple words: Wave nature of matter becomes visible when the particle's wavelength is similar to the size of obstacles or gaps it encounters, like atomic spacings in crystals.

📝 Teacher's Note: Emphasize that wave nature becomes apparent when wavelength is comparable to the size of the structures the particle interacts with.

🎯 Exam Tip: Wave effects are observable when λ ≈ size of diffracting object, typically on the order of atomic dimensions (10⁻¹⁰ m).