CBSE Class 12 Physics Dual Nature of Matter and Radiation Assignment

VERY SHORT ANSWER QUESTIONS (1 Mark)

Question 1. What is the rest mass of photon?
Answer: Zero.
In simple words: Photons always travel at the speed of light and do not have any rest mass.

Exam Tip: Remember that photons are pure packets of energy (quanta) with zero rest mass, but they possess relativistic mass, momentum, and energy.

Question 2. A good mirror reflects 80% of light incident on it. Which of the following is correct.
(a) Energy of each reflected photon decreases by 20%.
(b) Total no. of reflected photons decreases by 20%. Justify your answer.
Answer: (b) Total no. of reflected photons decreases by 20%
The energy of a photon depends only on its frequency (\(E = h\nu\)). Reflection from a mirror does not change the frequency of the photons. Therefore, energy of each reflected photon remains constant, and the 80% reflection means the number of reflected photons is 80% of the incident photons, so the total number of reflected photons decreases by 20%.
In simple words: The energy of individual photons does not change upon reflection, but fewer photons are reflected back.

Exam Tip: Intensity of light is proportional to the number of photons per unit area per second. Reflection loss reduces the number of photons, not their individual energies.

Question 3. Why in a photocell the cathode is coated with alkali metals?
Answer: Alkali metals have very low work functions, which means they can easily emit photoelectrons even when illuminated by visible light.
In simple words: Alkali metals need very little energy to release electrons, making them ideal for photoelectric cells.

Exam Tip: Mention low work function and sensitivity to visible light to score full marks.

Question 4. Name the phenomenon which shows quantum nature of electromagnetic radiation.
Answer: Photoelectric effect (and Compton effect).
In simple words: The photoelectric effect proves that light behaves like discrete packets (quanta) called photons.

Exam Tip: Photoelectric effect is the classic CBSE answer for quantum nature of radiation.

Question 5. Write Einstein's photoelectric equations and specify each term.
Answer: Einstein's photoelectric equation is given by:
\(K_{\max} = h\nu - \phi_0\)
Where:
\(K_{\max}\) = Maximum kinetic energy of emitted photoelectrons
\(h\) = Planck's constant
\(\nu\) = Frequency of incident radiation
\(\phi_0\) = Work function of the metal
In simple words: The maximum energy of the emitted electron equals the energy of the incoming photon minus the energy needed to escape the metal.

Exam Tip: Always define every single symbol used in the formula to avoid losing half marks.

Question 6. Which of the following radiations is more effective for electron emission from the surface of sodium?
(i) Microwave
(ii) Infrared
(iii) Ultraviolet
Answer: (iii) Ultraviolet
Ultraviolet radiation has a higher frequency and shorter wavelength, giving its photons energy well above the threshold frequency of sodium, compared to microwaves and infrared rays.
In simple words: Ultraviolet light has much higher energy than infrared or microwaves, making it capable of knocking out electrons.

Exam Tip: Higher frequency means higher photon energy (\(E = h\nu\)).

Question 7. A metal emits photoelectrons when red light falls on it. Will this metal emit photoelectrons when blue light falls on it?
Answer: Yes, it will emit photoelectrons.
Blue light has a higher frequency (and shorter wavelength) than red light. Since red light already causes photoelectric emission, blue light with greater energy per photon will easily cause emission with higher kinetic energy.
In simple words: Blue light has more energy than red light, so if red light works, blue light definitely will.

Question 8. Name any two phenomena which show the particle nature of radiation.
Answer: 1. Photoelectric effect
2. Compton effect
In simple words: Photoelectric effect and Compton effect both prove light acts like particles.

Exam Tip: Keep these two standard examples memorized for board exams.

Question 9. The photoelectric cut off voltage in a certain photoelectric experiment is 1.5V. What is the max kinetic energy of photoelectrons emitted?
Answer: Maximum kinetic energy is given by:
\(K_{\max} = eV_0\)
\(K_{\max} = 1.6 \times 10^{-19} \text{ C} \times 1.5 \text{ V} = 2.4 \times 10^{-19} \text{ J} = 1.5 \text{ eV}\)
In simple words: The maximum kinetic energy in electron-volts is numerically equal to the stopping potential in volts.

Exam Tip: 1 volt stopping potential equals 1 eV of maximum kinetic energy.

Question 10. What is the de-Broglie wavelength of a 3 kg object moving with a speed of 2m/s?
Answer: de-Broglie wavelength is given by:
\(\lambda = \frac{h}{mv}\)
Given: \(h = 6.63 \times 10^{-34} \text{ Js}\), \(m = 3 \text{ kg}\), \(v = 2 \text{ m/s}\)
\(\lambda = \frac{6.63 \times 10^{-34}}{3 \times 2} = \frac{6.63 \times 10^{-34}}{6} = 1.105 \times 10^{-34} \text{ m}\)
In simple words: Use the de-Broglie formula by dividing Planck's constant by momentum to get the wavelength.

Exam Tip: Always include proper SI units in your final numerical answer.

Question 11. What factors determine the maximum velocity of the photoelectrons from a surface?
Answer: 1. Frequency (or wavelength) of the incident radiation.
2. Work function of the metal surface.
In simple words: Faster electrons are emitted if the incoming light has higher frequency or if the metal has a lower work function.

Exam Tip: Note that intensity does not affect maximum velocity; intensity only affects the number of emitted electrons.

Question 12. What is the stopping potential applied to a photocell, in which electrons with a maximum kinetic energy of 5.6 eV are emitted.
Answer: Stopping potential \(V_0\) is given by:
\(eV_0 = K_{\max}\)
\(V_0 = \frac{K_{\max}}{e} = \frac{5.6 \text{ eV}}{e} = 5.6 \text{ V}\)
In simple words: The stopping potential is 5.6 V since 1 eV of energy corresponds to 1 V of stopping potential.

Exam Tip: Direct conversion: Energy in eV divided by electronic charge equals potential in volts.

Question 13. Work functions of caesium and lead are 2.14 eV and 4.25 eV respectively. Which of the two has a higher threshold wavelength?
Answer: Work function formula: \(\phi_0 = h\nu_0 = \frac{hc}{\lambda_0} \implies \lambda_0 \propto \frac{1}{\phi_0}\)
Since caesium has a lower work function than lead, caesium has a higher threshold wavelength.
In simple words: Metals with lower work functions require less energy, meaning they have longer threshold wavelengths.

Exam Tip: Threshold wavelength is inversely proportional to work function.

Question 14. What is the de-Broglie wavelength of a neutron at absolute temperature T K?
Answer: The kinetic energy of a thermal neutron at temperature \(T\) is \(E = \frac{3}{2}k_BT\).
de-Broglie wavelength is:
\(\lambda = \frac{h}{\sqrt{2m_nE}} = \frac{h}{\sqrt{2m_n \left(\frac{3}{2}k_BT\right)}} = \frac{h}{\sqrt{3m_nk_BT}}\)
Where \(k_B\) is Boltzmann's constant and \(m_n\) is the mass of the neutron.
In simple words: Substitute thermal kinetic energy into the de-Broglie wavelength formula.

Question 1. Write one similarity and one difference between matter wave and an electromagnetic wave.
Answer: Similarity: Both are independent of the charge of the particle and do not require any material medium for propagation.
Difference: Electromagnetic waves are transverse waves consisting of oscillating electric and magnetic fields, whereas matter waves are probability waves associated with moving particles and do not involve oscillating fields.
In simple words: Both travel without a medium, but EM waves are made of fields while matter waves represent moving particle probabilities.

Exam Tip: Clearly point out the nature of wave propagation and field oscillations.

Question 2. Does a photon have a de Broglie wavelength? Explain.
Answer: Yes, a photon has a de Broglie wavelength.
For a photon, momentum \(p = \frac{h}{\lambda}\), so its de Broglie wavelength is given by \(\lambda = \frac{h}{p} = \frac{h}{mc}\), which is identical to its wave wavelength.
In simple words: Photons possess momentum, so the de Broglie formula naturally applies to them as well.

Exam Tip: Mention that momentum links particle nature and wave nature for both matter and photons.

Question 3. A photon and an electron have energy 200 eV each. Which one of these has greater de-Broglie wavelength?
Answer: For a photon: Energy \(E = pc \implies p_{\text{photon}} = \frac{E}{c}\)
de-Broglie wavelength of photon: \(\lambda_{\text{photon}} = \frac{h}{p_{\text{photon}}} = \frac{hc}{E}\)
For an electron: Kinetic energy \(E = \frac{p_{\text{electron}}^2}{2m} \implies p_{\text{electron}} = \sqrt{2mE}\)
de-Broglie wavelength of electron: \(\lambda_{\text{electron}} = \frac{h}{\sqrt{2mE}}\)
Ratio: \(\frac{\lambda_{\text{photon}}}{\lambda_{\text{electron}}} = \frac{hc/E}{h/\sqrt{2mE}} = c \sqrt{\frac{2m}{E}}\)
Substituting values shows that for the same low energy of 200 eV, the electron has a much larger de Broglie wavelength than the photon.
In simple words: Calculating individual wavelengths shows the electron's wavelength is significantly greater than the photon's at 200 eV.

Exam Tip: Be careful with formulas: photon momentum uses \(E/c\), whereas massive particles use \(\sqrt{2mE}\).

Question 4. The work function of the following metal is given Na = 2.75 eV, K = 2.3 eV, Mo = 4.14 eV, Ni = 5.15 eV which of these metal will not give a photoelectric emission for radiation of wave length 3300 A° from a laser source placed at 1m away from the metal. What happens if the laser is brought nearer and placed 50 cm away.
Answer: Energy of incident photon corresponding to \(\lambda = 3300 \text{ \AA} = 3300 \times 10^{-10} \text{ m}\) is:
\(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{3300 \times 10^{-10} \times 1.6 \times 10^{-19}} \text{ eV}\)
\(E = \frac{19.89 \times 10^{-26}}{5.28 \times 10^{-28}} = 3.76 \text{ eV}\)
Metals with work functions greater than \(3.76 \text{ eV}\) will not emit photoelectrons. Therefore, Molybdenum (Mo = 4.14 eV) and Nickel (Ni = 5.15 eV) will not emit photoelectrons.
If the laser is brought closer to 50 cm, the intensity of light increases, increasing the number of photons, but the individual energy of each photon remains 3.76 eV. Hence, Mo and Ni will still not emit photoelectrons.
In simple words: Only metals with work functions lower than 3.76 eV emit electrons. Bringing the laser closer changes brightness, not photon energy.

Exam Tip: Distance changes intensity (number of photons), never the frequency or energy of individual photons.

Question 5. Name the experiment for which the followings graph, showing the variation of intensity of scattered electron with the angle of scattering, was obtained. Also name the important hypothesis that was confirmed by this experiment.
Answer: Experiment: Davisson and Germer experiment.
Hypothesis confirmed: De Broglie's hypothesis of wave nature of matter (electrons).
In simple words: The Davisson-Germer experiment proved that electrons can diffract like waves, confirming de Broglie's wave hypothesis.

Question 6. In a photoelectric effect experiment the graph between the stopping potential V and frequency of the incident radiation on two different metals P and Q are shown in Fig.
(i) Which of the two metals has greater value of work function?
(ii) Find maximum K.E. of electron emitted by light of frequency \(\nu = 8 \times 10^{14} \text{ Hz}\) for metal P.
Answer: (i) Metal Q has a greater threshold frequency (\(\nu_{0Q} > \nu_{0P}\)), so metal Q has a greater work function (\(\phi_0 = h\nu_0\)).
(ii) From the graph for metal P, threshold frequency \(\nu_{0P} = 4 \times 10^{14} \text{ Hz}\).
Maximum kinetic energy for \(\nu = 8 \times 10^{14} \text{ Hz}\) is:
\(K_{\max} = h(\nu - \nu_0) = 6.63 \times 10^{-34} \times (8 \times 10^{14} - 4 \times 10^{14})\)
\(K_{\max} = 6.63 \times 10^{-34} \times 4 \times 10^{14} = 2.65 \times 10^{-19} \text{ J} = 1.66 \text{ eV}\)
In simple words: Metal Q has a higher work function. For metal P, subtract threshold frequency from incident frequency and multiply by Planck's constant.

Exam Tip: The intercept on the frequency axis gives the threshold frequency.

Question 7. Do all the photons have same dynamic mass? If not, why?
Answer: No, all photons do not have the same dynamic mass.
Dynamic mass of a photon is given by \(m = \frac{h\nu}{c^2}\). Since different photons have different frequencies (\(\nu\)), their dynamic masses are different.
In simple words: Photon mass depends on frequency, so photons of different colors or energies have different masses.

Question 8. Why photoelectrons ejected from a metal surface have different kinetic energies although the frequency of incident photons are same?
Answer: Photoelectrons are emitted from different depths inside the metal surface. As they travel out, they suffer collisions and lose varying amounts of energy before escaping. Therefore, even though incident photons give the same initial energy, the emitted electrons emerge with different kinetic energies up to a maximum limit.
In simple words: Electrons deeper inside lose energy colliding with atoms on their way out, resulting in a range of kinetic energies.

Exam Tip: Mention internal collisions and energy losses below the surface.

Question 9. Find the ratio of de-Broglie wavelengths associated with two electrons 'A' and 'B' which are accelerated through 8V and 64 volts respectively.
Answer: de-Broglie wavelength for an electron accelerated through potential \(V\) is \(\lambda = \frac{12.27}{\sqrt{V}} \text{ \AA}\)
\(\frac{\lambda_A}{\lambda_B} = \sqrt{\frac{V_B}{V_A}} = \sqrt{\frac{64}{8}} = \sqrt{8} = 2\sqrt{2}\)
In simple words: Use the inverse square root relation between wavelength and accelerating potential.

Exam Tip: Use the shortcut formula \(\lambda = \frac{12.27}{\sqrt{V}} \text{ \AA}\) for quick calculations involving electrons.

Question 10. The photoelectric current at distances \(r_1\) and \(r_2\) of light source from photoelectric cell are \(I_1\) and \(I_2\) respectively. Find the value of \(\frac{I_2}{I_1}\).
Answer: Intensity of light is inversely proportional to the square of the distance (\(I \propto \frac{1}{r^2}\)).
Photoelectric current is directly proportional to intensity.
Therefore, \(\frac{I_2}{I_1} = \left(\frac{r_1}{r_2}\right)^2\)
In simple words: Intensity and current drop with the square of the distance from the light source.

Exam Tip: Remember inverse square law for point sources in optics and modern physics.

Question 11. How does the maximum kinetic energy of emitted electrons vary with the increase in work function of metals?
Answer: From Einstein's equation \(K_{\max} = h\nu - \phi_0\), maximum kinetic energy decreases linearly with an increase in work function \(\phi_0\).
In simple words: Higher work function means the metal holds onto electrons tighter, leaving them with less kinetic energy.

Question 1. Explain the working of a photocell? Give its two uses.
Answer: Working: A photocell is a device based on the photoelectric effect that converts light energy into electrical energy. When light of suitable frequency falls on the photosensitive cathode, photoelectrons are emitted and attracted toward the positive anode, producing a photoelectric current in the external circuit.
Two uses: 1. In television camera scanning and sound reproduction in motion pictures.
2. In automatic fire alarms and burglar alarms.
In simple words: Photocells turn light into electricity when photons knock electrons from a cathode to an anode. Used in alarms and cinema sound tracks.

Exam Tip: Clearly write both the principle/working mechanism and practical applications.

Question 2. Find the de Broglie wavelength associated with an electron accelerated through a potential difference V.
Answer: Kinetic energy of an electron accelerated through potential \(V\) is \(K = eV\).
Momentum \(p = \sqrt{2mK} = \sqrt{2meV}\)
de-Broglie wavelength \(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}\)
Substituting values of constants (\(h = 6.63 \times 10^{-34} \text{ Js}\), \(m = 9.1 \times 10^{-31} \text{ kg}\), \(e = 1.6 \times 10^{-19} \text{ C}\)):
\(\lambda = \frac{12.27}{\sqrt{V}} \text{ \AA}\)
In simple words: Derive momentum from kinetic energy and substitute it into de Broglie's formula.

Exam Tip: This is a mandatory 3-mark derivation. Practice every step carefully.

Question 3. What is Einstein's explanation of photo electric effect? Explain the laws of photo electric emission on the basis of quantum nature of light.
Answer: Einstein explained that light travels in discrete packets of energy called photons (\(E = h\nu\)). When a photon collides with an electron, it transfers its entire energy to the single electron instantaneously.
Laws explained:
1. Threshold frequency: Emission occurs only if photon energy exceeds work function (\(h\nu \ge \phi_0\)).
2. Kinetic energy independence of intensity: Increasing intensity increases the number of photons, not their energy, so kinetic energy depends only on frequency.
3. Instantaneous emission: Photon-electron collision is direct and instantaneous without time lag.
In simple words: One photon hits one electron instantly. Higher frequency gives more energy, while higher intensity gives more electrons.

Exam Tip: Structure your answer with Einstein's photon hypothesis followed by point-by-point explanation of the laws.

Question 4. If kinetic energy of thermal neutron is \(\frac{3}{2}kT\) then show that de-Broglie wavelength of waves associated with a thermal neutron of mass m at temperature T Kelvin is \(\frac{h}{\sqrt{3mkT}}\) where k is boltz mann constant.
Answer: Given kinetic energy \(E = \frac{3}{2}kT\).
Momentum of neutron \(p = \sqrt{2mE} = \sqrt{2m \left(\frac{3}{2}kT\right)} = \sqrt{3mkT}\)
de-Broglie wavelength \(\lambda = \frac{h}{p} = \frac{h}{\sqrt{3mkT}}\)
In simple words: Substitute thermal kinetic energy into the momentum expression and put it into the de Broglie equation.

Exam Tip: Simple substitution makes this an easy 3-mark derivation.

Question 5. Explain Davisson and Germer experiment to verify the wave nature of electrons.
Answer: Setup: An electron gun emits electrons accelerated by potential \(V\). The electron beam is made to fall on a single nickel crystal, and the scattered electrons are detected by a Faraday cup detector at various scattering angles \(\theta\).
Observation: At an accelerating voltage of \(54 \text{ V}\) and scattering angle \(\theta = 50^\circ\), a sharp peak in electron scattering intensity was observed, which corresponds to Bragg's law of X-ray diffraction.
Conclusion: This confirmed that electrons exhibit wave-like behavior with wavelength matching de Broglie's theoretical prediction (\(\lambda = 1.66 \text{ \AA}\)).
In simple words: Firing electrons at a nickel crystal created a diffraction pattern, proving electrons act like waves.

Exam Tip: Mention 54 V, 50 degrees, and nickel crystal as key experimental parameters.

Question 6. Explain the effect of increase of (i) frequency (ii) intensity of the incident radiation on photo electrons emitted by a metal.
Answer: (i) Increase in frequency: Increases the energy of incident photons, which increases the maximum kinetic energy and maximum velocity of emitted photoelectrons. The stopping potential also increases.
(ii) Increase in intensity: Increases the number of photons striking the surface per second, which increases the photoelectric current (number of emitted electrons). However, kinetic energy and stopping potential remain unchanged.
In simple words: Frequency changes electron speed and energy; intensity changes the number of electrons.

Exam Tip: Never confuse intensity with frequency in photoelectric effect questions.

Question 7. X-rays of wavelength \(\lambda\) fall on a photo sensitive surface emitting electrons. Assuming that the work function of the surface can be neglected, prove that the de-Broglie wavelength of electrons emitted will be \(\sqrt{\frac{h\lambda}{2mc}}\).
Answer: Energy of incident X-ray photon \(E = \frac{hc}{\lambda}\)
Since work function is neglected, maximum kinetic energy of emitted electron is \(K = \frac{hc}{\lambda}\)
Momentum of emitted electron \(p = \sqrt{2mK} = \sqrt{2m \left(\frac{hc}{\lambda}\right)} = \sqrt{\frac{2mhc}{\lambda}}\)
de-Broglie wavelength of electron \(\lambda_e = \frac{h}{p} = \frac{h}{\sqrt{\frac{2mhc}{\lambda}}} = \sqrt{\frac{h^2 \lambda}{2mhc}} = \sqrt{\frac{h\lambda}{2mc}}\)
In simple words: Equate electron kinetic energy to photon energy, find momentum, and substitute into the de Broglie formula.

Question 8. A particle of mass M at rest decays into two particles of masses \(m_1\) and \(m_2\) having velocities \(V_1\) and \(V_2\) respectively. Find the ratio of de-broglie Wavelengths of the two particles.
Answer: By the law of conservation of momentum:
\(m_1V_1 + m_2V_2 = 0 \implies |m_1V_1| = |m_2V_2|\)
Since magnitudes of momenta are equal (\(p_1 = p_2\)):
\(\lambda_1 = \frac{h}{p_1}\) and \(\lambda_2 = \frac{h}{p_2}\)
Therefore, \(\frac{\lambda_1}{\lambda_2} = \frac{p_2}{p_1} = 1 : 1\)
In simple words: Conservation of momentum ensures both decay fragments have equal and opposite momenta, leading to identical de Broglie wavelengths.

Exam Tip: Equal momenta always result in a 1:1 de Broglie wavelength ratio.

NUMERICALS

Question 1. Ultraviolet light of wavelength \(350 \text{ nm}\) and intensity \(1 \text{ W/m}^2\) is directed at a potassium surface having work function \(2.2 \text{ eV}\).
(i) Find the maximum kinetic energy of the photoelectron.
(ii) If 0.5 percent of the incident photons produce photoelectric effect, how many photoelectrons per second are emitted from the potassium surface that has an area \(1 \text{ cm}^2\).
Answer: (i) Energy of incident photon:
\(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{350 \times 10^{-9}} = 5.68 \times 10^{-19} \text{ J} = \frac{5.68 \times 10^{-19}}{1.6 \times 10^{-19}} = 3.55 \text{ eV}\)
Maximum kinetic energy \(K_{\max} = E - \phi_0 = 3.55 \text{ eV} - 2.2 \text{ eV} = 1.35 \text{ eV}\)
(ii) Intensity \(I = 1 \text{ W/m}^2\), Area \(A = 1 \text{ cm}^2 = 10^{-4} \text{ m}^2\)
Total incident power \(P = I \times A = 1 \times 10^{-4} \text{ W}\)
Energy per photon \(E = 5.68 \times 10^{-19} \text{ J}\)
Number of incident photons per second \(n = \frac{P}{E} = \frac{10^{-4}}{5.68 \times 10^{-19}} = 1.76 \times 10^{15} \text{ s}^{-1}\)
Number of photoelectrons emitted per second = \(0.5\% \text{ of } n = 0.005 \times 1.76 \times 10^{15} = 8.8 \times 10^{12} \text{ s}^{-1}\)
In simple words: Calculate photon energy first, find max KE by subtracting work function, then use total power to find electron emission rate.

Question 2. A metal surface illuminated by \(8.5 \times 10^{14} \text{ Hz}\) light emits electrons whose maximum energy is \(0.52 \text{ eV}\) the same surface is illuminated by \(12.0 \times 10^{14} \text{ Hz}\) light emits elections whose maximum energy is \(1.97\text{ eV}\). From these data find work function of the surface and value of Planck's constant. [Work Function = 3eV]
Answer: Using Einstein's photoelectric equations:
\(K_1 = h\nu_1 - \phi_0 \implies 0.52 \times 1.6 \times 10^{-19} = h(8.5 \times 10^{14}) - \phi_0\) --- (1)
\(K_2 = h\nu_2 - \phi_0 \implies 1.97 \times 1.6 \times 10^{-19} = h(12.0 \times 10^{14}) - \phi_0\) --- (2)
Subtracting (1) from (2):
\((1.97 - 0.52) \times 1.6 \times 10^{-19} = h(12.0 - 8.5) \times 10^{14}\)
\(1.45 \times 1.6 \times 10^{-19} = h \times 3.5 \times 10^{14}\)
\(h = \frac{2.32 \times 10^{-19}}{3.5 \times 10^{14}} = 6.63 \times 10^{-34} \text{ Js}\)
Substituting \(h\) into equation (1) to find work function \(\phi_0\):
\(\phi_0 = h\nu_1 - K_1 = (6.63 \times 10^{-34} \times 8.5 \times 10^{14}) - (0.52 \times 1.6 \times 10^{-19})\)
\(\phi_0 = 5.635 \times 10^{-19} - 0.832 \times 10^{-19} = 4.80 \times 10^{-19} \text{ J} = 3.0 \text{ eV}\)
In simple words: Set up two simultaneous equations for the two frequencies and solve for Planck's constant and work function.

Exam Tip: This two-frequency method is a standard 5-mark numerical type in board exams.

Question 3. An electron and photon each have a wavelength of \(0.2 \text{ nm}\). Calculate their momentum and energy.
(i) \(3.3 \times 10^{-24} \text{ kgm/s}\)
(ii) \(6.2 \text{ keV} \text{ for photon}\)
(iii) \(38 \text{ eV} \text{ for electron}\)
Answer: Given wavelength \(\lambda = 0.2 \text{ nm} = 0.2 \times 10^{-9} \text{ m}\)
Momentum for both electron and photon: \(p = \frac{h}{\lambda} = \frac{6.63 \times 10^{-34}}{0.2 \times 10^{-9}} = 3.315 \times 10^{-24} \text{ kg m/s}\)
(ii) Energy of photon: \(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{0.2 \times 10^{-9}} = 9.945 \times 10^{-16} \text{ J} = 6.21 \text{ keV}\)
(iii) Energy of electron: \(E = \frac{p^2}{2m} = \frac{(3.315 \times 10^{-24})^2}{2 \times 9.1 \times 10^{-31}} = 6.03 \times 10^{-18} \text{ J} = 37.7 \text{ eV} \approx 38 \text{ eV}\)
In simple words: Both share the same momentum from their wavelength, but use different energy formulas for photons and electrons.

Exam Tip: Pay attention to energy formulas: \(hc/\lambda\) for photons and \(p^2/2m\) for massive particles.

Question 4. What is the (i) Speed (ii) Momentum (iii) de-Broglie wavelength of an electron having kinetic energy of 120eV?
[Ans. : (a) \(6.5 \times 10^6 \text{ m/s}\); (b) \(5.92 \times 10^{-24} \text{ kg m/s}\); (c) \(0.112 \text{ nm}\)]
Answer: Given Kinetic Energy \(K = 120 \text{ eV} = 120 \times 1.6 \times 10^{-19} \text{ J} = 1.92 \times 10^{-17} \text{ J}\)
(i) Speed \(v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2 \times 1.92 \times 10^{-17}}{9.1 \times 10^{-31}}} = 6.49 \times 10^6 \text{ m/s}\)
(ii) Momentum \(p = mv = 9.1 \times 10^{-31} \times 6.49 \times 10^6 = 5.91 \times 10^{-24} \text{ kg m/s}\)
(iii) de-Broglie wavelength \(\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{5.91 \times 10^{-24}} = 1.12 \times 10^{-10} \text{ m} = 0.112 \text{ nm}\)
In simple words: Find speed from kinetic energy, calculate momentum, and then compute the de Broglie wavelength.

Question 5. If the frequency of incident light in photoelectric experiment is doubled then does the stopping potential become double or more than double, justify? (More than double)
Answer: Einstein's equation: \(eV_0 = h\nu - \phi_0 \implies V_0 = \frac{h\nu}{e} - \frac{\phi_0}{e}\)
When frequency \(\nu\) is doubled to \(2\nu\), the new stopping potential is \(V_0' = \frac{2h\nu}{e} - \frac{\phi_0}{e} = 2\left(\frac{h\nu}{e}\right) - \frac{\phi_0}{e}\)
Clearly, \(V_0' > 2V_0\), so the stopping potential becomes more than double.
In simple words: Doubling frequency doubles the photon energy term, but the work function subtraction makes the new stopping potential more than double.

Exam Tip: Always account for the constant work function term when scaling frequencies.

Question 6. A proton is accelerated through a potential difference V. Find the percentage increase or decrease in its deBroglie wavelength if potential difference is increased by 21%. (9.1%)
Answer: de-Broglie wavelength of a proton is \(\lambda = \frac{h}{\sqrt{2mqV}} \implies \lambda \propto \frac{1}{\sqrt{V}}\)
New potential \(V' = V + 0.21V = 1.21V\)
New wavelength \(\lambda' = \frac{h}{\sqrt{2mq(1.21V)}} = \frac{\lambda}{\sqrt{1.21}} = \frac{\lambda}{1.1}\)
Percentage decrease in wavelength = \(\frac{\lambda - \lambda'}{\lambda} \times 100 = \frac{\lambda - \frac{\lambda}{1.1}}{\lambda} \times 100 = \left(1 - \frac{1}{1.1}\right) \times 100 = 9.09\% \approx 9.1\%\)
In simple words: Use proportionality with potential, calculate the new wavelength factor, and find the percentage change.

Exam Tip: \(\sqrt{1.21} = 1.1\) makes this percentage calculation straightforward.

Question 7. For what Kinetic energy of a neutron will the associated de Broglie wavelength be \(5.6 \times 10^{-10} \text{ m}\)?
Answer: de-Broglie wavelength formula: \(\lambda = \frac{h}{\sqrt{2m_nK}} \implies K = \frac{h^2}{\lambda^2 2m_n}\)
Given: \(\lambda = 5.6 \times 10^{-10} \text{ m}\), \(h = 6.625 \times 10^{-34} \text{ Js}\), \(m_n = 1.675 \times 10^{-27} \text{ kg}\)
\(K = \frac{(6.625 \times 10^{-34})^2}{(5.6 \times 10^{-10})^2 \times 2 \times 1.675 \times 10^{-27}}\)
\(K = \frac{43.9 \times 10^{-68}}{31.36 \times 10^{-20} \times 3.35 \times 10^{-27}} = \frac{43.9 \times 10^{-68}}{105.06 \times 10^{-47}} = 4.17 \times 10^{-21} \text{ J} = 0.026 \text{ eV}\)
In simple words: Rearrange the de Broglie formula to solve directly for kinetic energy.

Exam Tip: Substitute neutron mass carefully instead of electron mass.

Question 8. A nucleus of mass M initially at rest splits into two fragments of masses \(\frac{M}{3}\) and \(\frac{2M}{3}\). Find the ratio of de Broglie wavelength of the fragments.
Answer: Following the law of conservation of momentum:
\(\frac{M}{3}V_1 + \frac{2M}{3}V_2 = 0 \implies \left|\frac{M}{3}V_1\right| = \left|\frac{2M}{3}V_2\right|\)
Since momenta are equal in magnitude (\(p_1 = p_2\)):
\(\lambda = \frac{h}{p} \implies \frac{\lambda_1}{\lambda_2} = \frac{p_2}{p_1} = 1 : 1\)
In simple words: Conservation of momentum ensures both fragments carry equal and opposite momentum, giving a 1:1 de Broglie wavelength ratio.

Question 9. An electron and a proton are possessing same amount of K.E., which of the two have greater de-Broglie, wavelength? Justify your answer.
Answer: Kinetic energy: \(E_e = \frac{1}{2}m_ev_e^2\) and \(E_p = \frac{1}{2}m_pv_p^2\)
Momentum: \(p_e = \sqrt{2m_eE_e}\) and \(p_p = \sqrt{2m_pE_p}\)
Since \(E_e = E_p\), the ratio of de-Broglie wavelengths is:
\(\frac{\lambda_e}{\lambda_p} = \frac{p_p}{p_e} = \sqrt{\frac{m_p}{m_e}}\)
Since mass of proton is greater than mass of electron (\(m_p > m_e\)), \(\frac{\lambda_e}{\lambda_p} > 1 \implies \lambda_e > \lambda_p\).
Therefore, the electron has a greater de-Broglie wavelength.
In simple words: For equal kinetic energy, the lighter particle (electron) has a larger de Broglie wavelength.

Exam Tip: Remember the inverse mass relationship under equal kinetic energy conditions.