CBSE Class 12 Physics Dual Nature of Matter Assignment

Important Questions for NCERT Class 12 Physics Dual Nature Of Matter And Radiation 

Question. In a photoelectric effect experiment, for radiation with frequency υ₀ with hυ₀ = 8eV, electrons are emitted with energy 2 eV. What is the energy of the electrons emitted for incoming radiation of frequency 1.25 υ₀ ?
(a) 1 eV
(b) 3.25 eV
(c) 4 eV
(d) 9.25 eV.

Answer: C

Question. Einstein’s photoelectric equation is E_k = h_n - Φ . In this equation Ek refers to
(a) kinetic energy of all the emitted electrons
(b) mean kinetic energy of emitted electrons
(c) maximum kinetic energy of emitted electrons
(d) minimum kinetic energy of emitted electrons

Answer: C

Question. Photoelectric effect is the phenomenon in which
(a) photons come out of a metal when it is hit by a beam of electrons.
(b) photons come out of the nucleus of an atom under the action of an electric field.
(c) electrons come out of a metal with a constant velocity which depends on the frequency and intensity of incident light wave.
(d) electons come out of a metal with different velocities not greater than a certain value which depends only on the frequency of the incident light wave and not on its intensity.

Answer: D

Question. A photoelectric cell is a device which
(a) converts light into electricity
(b) converts electricity into light
(c) stores light
(d) stores electricity

Answer: A

Question. If E₁, E₂, E₃ are the respective kinetic energies of an electron, an alpha-particle and a proton, each having the same de-Broglie wavelength, then
(a) E₁ > E₃ > E₂
(b) E₂ > E₃ > E₁
(c) E₁ > E₂ > E₃
(d) E₁ = E₂ = E₃

Answer: A

Question. When the speed of electrons increase, then the value of its specific charge
(a) increases
(b) decreases
(c) ramains unchanged
(d) increases upto some velocity and then begins to decrease

Answer: B

Question. In an electron gun the control grid is given a negative potential relative to cathode in order to
(a) decelerate electrons
(b) repel electrons and thus to control the number of electrons passing through it
(c) to select electrons of same velocity and to converge them along the axis.
(d) to decrease the kinetic energy of electrons

Answer: B

Question. X-rays are produced in X-ray tube operating at a given accelerating voltage. The wavelength of the continuous X-rays has values from
(a) 0 to ∞
(b) λ_min to ∞, where λ_min > 0
(c) 0 to λ_max, where λ_max < ∞
(d) λ_min to λ_max, where 0 < λ_min <λ_max < ∞

Answer: B

Question. As intensity of incident light increases
(a) photoelectric current increase
(b) K.E. of emitted photoelectron increases
(c) photoelectric current decreases
(d) K.E. of emitted photoelectrons decreases

Answer: A

Question. Which of the following shows par.ticle nature of light?
(a) Refraction
(b) Interference
(c) Polarization
(d) Photoelectric effect

Answer: D

Question. The ratio of de-Broglie wavelengths of proton and a -particle having same kinetic energy is
(a) √2 : 1
(b) 2 √2 : 1
(c) 2 : 1
(d) 4 : 1

Answer: C

Question. White X-rays are called white due to the fact that
(a) they are electromagnetic radiations having nature same as that of white light.
(b) they are produced most abundantly in X ray tubes.
(c) they have a continuous wavelength range.
(d) they can be converted to visible light using coated screens and photographic plates are affected by them just like light.

Answer: C

Question. In case of electrons and photons having the same wavelength. What is same for them?
(a) Energy
(b) Velocity
(c) Momentum
(d) Angular momentum

Answer: C

Question. The energy of a photon of wavelength λ is
(a) hc λ
(b) hc/λ
(c) λ/hc
(d) λh/c

Answer: B

Question. In the Davisson and Germer experiment, the velocity of electrons emitted from the electron gun can be increased by
(a) increasing the potential difference between the anode and filament
(b) increasing the filament current
(c) decreasing the filament current
(d) decreasing the potential difference between the anode and filament

Answer: A

Question. In a photoelectric experiment the stopping potential for the incident light of wavelength 4000Å is 2 volt. If the wavelength be changed to 3000 Å, the stopping potential will be
(a) 2 V
(b) zero
(c) less than 2 V
(d) more than 2 V

Answer: D

Question. A proton and a-particle are accelerated through the same potential difference. The ratio of their de-Broglie wavelength will be
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 2 √2 : 1

Answer: D

Question. The stopping potential (V₀) versus frequency (v) plot of a substance is shown in figure, the threshold wavelength is 

(a) 5 × 1014m
(b) 6000 Å
(c) 5000 Å
(d) Cannot be estimated from given data

Answer: B

Question. The de-Broglie wavelength of a proton (mass = 1.6 × 10^–27 kg) accelerated through a potential difference of 1 kV is
(a) 600 A
(b) 0.9 × 10^–12m
(c) 7 Å
(d) 0.9 nm.

Answer: B

Question. An electron is accelerated by a p.d. of 1000 V. Its velocity will be
(a) 3.78 × 10⁷ m/s
(b) 1.89 × 10⁷ m/s
(c) 5.67 × 10⁷ m/s
(d) 0.95 × 10⁷ m/s

Answer: B

Question. If a photon has velocity c and frequency u, then which of the following represents its wavelength?
(a) hν/c²
(b) hν
(c) hν/E
(d) hν/c

Answer :  C

Question. Momentum of photon of wavelength l is
(a) hν/c
(b) zero
(c) hν/C²
(d) hν/c

Answer :  A

Question. The wavelength of a 1 keV photon is 1.24 × 10^–9 m.
What is the frequency of 1 MeV photon?
(a) 1.24 × 10¹⁵
(b) 2.4 × 10²⁰
(c) 1.24 × 10¹⁸
(d) 2.4 × 10²³ 

Answer :  B

Question. A radio transmitter operates at a frequency 880 kHz and a power of 10 kW. The number of photons emitted per second is
(a) 1.72 × 10³¹
(b) 1.327 × 10²⁵
(c) 1.327 × 10³⁷
(d) 1.327 × 10⁴⁵ 

Answer :  A

Question. The momentum of a photon of an electromagnetic radiation is 3.3 ×10^–29 kg m s^–1. What is the frequency of the associated waves? [h = 6.6 × 10^–34 J s ; c = 3 × 10⁸ m s^–1]
(a) 1.5 × 10¹³ Hz
(b) 7.5 × 10¹² Hz
(c) 6 × 10³ Hz
(d) 3 × 10³ Hz 

Answer :  A

Question. The energy of a photon of wavelength l is
(a) hcλ (b) hc/λ
(c) λ/hc
(d) λh/c

Answer :  B

Question. An electron is accelerated from rest through a potential difference of V volt. If the de Broglie wavelength of the electron is 1.227 × 10–2 nm, the potential difference is
(a) 10 V
(b) 10² V
(c) 10³ V
(d) 10⁴ V 

Answer :  A

Question. An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is, (nearly) (me = 9 × 10^–31 kg)
(a) 12.2 nm
(b) 12.2 × 10^–13 m
(c) 12.2 × 10^–12 m
(d) 12.2 × 10^–14 m

Answer :  C

Question. Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de Broglie wavelength of the emitted electron is
(a) ≥ 2.8 × 10^–9 m
(b) ≤ 2.8 × 10^–12 m
(c) < 2.8 × 10^–10 m
(d) < 2.8 × 10^–9 m

Answer :  A

Question. If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de Broglie wavelength of the particle is
(a) 25
(b) 75
(c) 60
(d) 50

Answer :  B

Question. An a-particle moves in a circular path of radius 0.83 cm in the presence of a magnetic field of 0.25 Wb/m2. The de Broglie wavelength associated with the particle will be
(a) 1 Å
(b) 0.1 Å
(c) 10 Å
(d) 0.01 Å 

Answer :  D

DUAL NATURE OF MATTER AND RADIATION

TOPICS TO BE COVERED

Dual nature of radiation, Photoelectric effect Hertz and Lenard's observations; Einstein's photoelectrical equation, Particle nature of light.

Matter waves-wave nature of particles, de-Broglie relation Davisson Germer experiment.

KEY POINTS

- Light consists of individual photons whose energies are proportional to their frequencies.
- A photon is a quantum of electromagnetic energy:
Energy of photon: \(E = h\nu = \frac{hc}{\lambda}\)
Momentum of a photon: \(p = \frac{h\nu}{c} = \frac{h}{\lambda}\)
Dynamic mass of photon: \(m = \frac{h\nu}{c^2} = \frac{h}{c\lambda}\)
Rest mass of a photon is zero.
- Photoelectric effect: Photon of incident light energy interacts with a single electron and if energy of photon is equal to or greater than work function, the electron is emitted.
- Max. kinetic energy of emitted electron = \(h(\nu - \nu_0)\) Here \(\nu_0\) is the frequency below which no photoelectron is emitted and is called threshold frequency.
- A moving body behaves in a certain way as though it has a wave nature having wavelength, \(\lambda = \frac{h}{mv}\).

QUESTIONS

VERY SHORT ANSWER QUESTIONS (1 Mark)

Question 1. What is the rest mass of photon?
Answer: Zero.
In simple words: Photons are packets of energy that always travel at the speed of light and have no rest mass.

Exam Tip: Always remember that rest mass of photon \(m_0 = 0\), but it possesses dynamic mass given by \(m = \frac{h}{\lambda c}\).

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%.
Answer: (b) Total no. of reflected photons decreases by 20%.
Justification: Reflection of light is a particle phenomenon where each photon is either completely reflected or absorbed. When 80% of light is reflected, it means 80% of the total incident photons are reflected back, so the total number of reflected photons decreases by 20%. The energy and frequency of each individual reflected photon remain unchanged.
In simple words: The energy of a photon depends on its frequency, which does not change upon reflection. Only the number of photons decreases.

Exam Tip: Energy of a photon is \(E = h\nu\). Since frequency \(\nu\) does not change during reflection, the energy of each photon remains constant.

Question 3. Why in a photocell the cathode is coated with alkali metals?
Answer: Alkali metals have very low work functions and low ionization energies, 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 photocells.

Exam Tip: Mention low work function as the primary reason for using alkali metals.

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 of energy (particles/quanta) rather than continuous waves.

Exam Tip: Photoelectric effect is the most standard answer expected in exams for quantum nature.

Question 5. Write Einstein's photoelectric equations and specify each term.
Answer: Einstein's photoelectric equation is:
\(K_{\max} = h\nu - \Phi_0\) or \(eV_0 = h\nu - h\nu_0\)
Where:
- \(K_{\max}\) = Maximum kinetic energy of emitted photoelectrons
- \(h\nu\) = Energy of the incident photon
- \(\Phi_0\) (or \(h\nu_0\)) = Work function of the metal
- \(V_0\) = Stopping potential
- \(\nu_0\) = Threshold frequency
In simple words: The energy of incoming light goes into releasing the electron (work function) and giving it kinetic energy.

Exam Tip: Clearly define every symbol used in the equation to avoid losing 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 much higher frequency and shorter wavelength compared to microwaves and infrared, giving individual photons sufficient energy to exceed the work function of sodium.
In simple words: Higher frequency radiation has more energy per photon, making it effective for knocking out electrons.

Exam Tip: Frequency is directly proportional to photon energy (\(E = h\nu\)), so high-frequency waves like UV rays are most effective.

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, the metal will emit photoelectrons.
Reason: Blue light has a higher frequency (and shorter wavelength) than red light (\(\nu_{\text{blue}} > \nu_{\text{red}}\)). Since red light is already able to cause emission, blue light with higher energy will definitely exceed the threshold frequency.
In simple words: Blue light has more energy than red light, so it easily causes electron emission.

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 that light acts like particles.

Exam Tip: Memorize these two standard quantum phenomena.

Question 9. The photoelectric cut-off voltage in a certain photoelectric experiment is 1.5 V. What is the max kinetic energy of photoelectrons emitted?
Answer: Maximum kinetic energy is given by:
\(K_{\max} = eV_0\)
\(K_{\max} = 1 \, \text{e} \times 1.5 \, \text{V} = 1.5 \, \text{eV} = 1.5 \times 1.6 \times 10^{-19} \, \text{J} = 2.4 \times 10^{-19} \, \text{J}\)
In simple words: Kinetic energy in electron-volts is numerically equal to the stopping potential in volts.

Exam Tip: \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\). Keep your answers in eV unless Joules are specifically requested.

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

Exam Tip: Check standard values of Planck's constant \(h = 6.63 \times 10^{-34} \, \text{Js}\) during calculations.

Question 11. What factors determine the maximum velocity of the photoelectrons from a surface?
Answer: The maximum velocity of photoelectrons depends on:
1. Frequency (or wavelength) of the incident radiation.
2. Work function (or threshold frequency) of the emitter metal surface.
In simple words: Higher frequency light and lower work function metals produce faster electrons.

Exam Tip: Intensity does not affect maximum velocity; it only affects the number of electrons emitted.

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: Given \(K_{\max} = 5.6 \, \text{eV}\).
Stopping potential \(V_0\) is given by:
\(eV_0 = K_{\max}\)
\(V_0 = \frac{5.6 \, \text{eV}}{e} = 5.6 \, \text{V}\)
In simple words: The stopping potential in volts is numerically equal to the max kinetic energy in eV.

Exam Tip: Direct conversion applies when energy is expressed in eV.

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 (2.14 eV) compared to lead (4.25 eV), it has a higher threshold wavelength.
In simple words: Work function is inversely proportional to threshold wavelength, so lower work function means higher threshold wavelength.

Exam Tip: Always remember \(\lambda_0 = \frac{hc}{\Phi_0}\).

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 absolute temperature \(T\) is given by \(E = \frac{3}{2} k_B T\).
De-Broglie wavelength is:
\(\lambda = \frac{h}{\sqrt{2m_n E}} = \frac{h}{\sqrt{2m_n \left(\frac{3}{2} k_B T\right)}} = \frac{h}{\sqrt{3m_n k_B T}}\)
Where \(k_B\) is Boltzmann's constant and \(m_n\) is the mass of the neutron.
In simple words: Substitute thermal kinetic energy \(\frac{3}{2} k_B T\) into the de-Broglie formula.

Exam Tip: This is a very important derivation frequently asked in exams.

SHORT ANSWER QUESTIONS (2 Marks)

Question 1. Write one similarity and one difference between matter wave and an electromagnetic wave.
Answer: Similarity: Both waves are independent of the charge of the particle and do not require any mechanical 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 electric or magnetic field oscillations.
In simple words: Both travel in a vacuum, but EM waves are made of fields while matter waves represent moving particles.

Exam Tip: Focus on the physical nature of the waves for clear distinction.

Question 2. Does a photon have a de Broglie wavelength? Explain.
Answer: No, because de-Broglie wavelength is defined for material particles with rest mass (\(\lambda = \frac{h}{mv}\)). For a photon, mass is zero and its wavelength is simply its electromagnetic wavelength given by \(\lambda = \frac{c}{\nu}\). Alternatively, using relativistic momentum: \(m = \frac{E}{c^2} = \frac{h\nu}{c^2}\), the concept of de-Broglie wavelength reduces to standard photon wavelength \(\frac{h}{mc} = \frac{h}{(h\nu/c^2)c} = \frac{c}{\nu}\).
In simple words: Photons already have a wave wavelength associated with them, so the term de-Broglie wavelength is specifically reserved for material particles.

Exam Tip: Be precise about definitions when dealing with photons versus massive particles.

Question 3. A photon and an electron have energy 200 eV each. Which one of these has greater de-Broglie wavelength?
Answer: For the electron:
\(\lambda_e = \frac{h}{\sqrt{2m_e E}}\)
For the photon, energy is \(E = pc \implies p = \frac{E}{c}\), so wavelength is:
\(\lambda_p = \frac{h}{p} = \frac{hc}{E}\)
Comparing their expressions for 200 eV energy, the electron has a much larger de-Broglie wavelength than the photon because of the mass factor in the denominator of the electron formula.
In simple words: The electron will have a larger wavelength due to its mass scaling in the momentum equation.

Exam Tip: Always derive or use respective formulas for particle vs photon wavelength.

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 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 eV will not emit photoelectrons. Therefore, Mo (4.14 eV) and Ni (5.15 eV) will not give photoelectric emission.
If the laser is brought closer to 50 cm, the intensity of light increases, but the energy of individual photons remains 3.76 eV. Hence, there is still no emission from Mo and Ni because emission depends on photon energy (frequency), not intensity.
In simple words: Photon energy is 3.76 eV. Metals requiring more energy (Mo and Ni) will not emit electrons, and changing distance only changes brightness, not photon energy.

Exam Tip: Intensity affects the number of photons, but energy per photon depends solely on wavelength/frequency.

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 name: Davisson-Germer experiment.
Hypothesis confirmed: de-Broglie hypothesis of wave nature of moving particles (electrons).
In simple words: Davisson-Germer experiment proved that electrons can act like waves and show diffraction patterns.

Exam Tip: Remember that Davisson-Germer proved electron wave nature using a nickel crystal target.

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 P has a higher threshold frequency (\(\nu_{0P} > \nu_{0Q}\)), and since work function \(\Phi_0 = h\nu_0\), metal P has a greater work function.
(ii) From the graph for metal P, threshold frequency \(\nu_{0P} = 4 \times 10^{14} \, \text{Hz}\).
Max kinetic energy = \(h(\nu - \nu_0)\)
\(K_{\max} = h(8 \times 10^{14} - 4 \times 10^{14}) = h(4 \times 10^{14}) \, \text{J}\)
In simple words: Higher threshold frequency means a higher work function, and kinetic energy is calculated from the frequency difference above threshold.

Exam Tip: Intercept on the frequency axis gives the threshold frequency.

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

Exam Tip: Relate dynamic mass directly to frequency using Einstein-Planck relations.

Question 8. Why photoelectrons ejected from a metal surface have different kinetic energies although the frequency of incident photons are same?
Answer: Electrons inside a metal are at different depths and encounter varying levels of collisions as they absorb a photon and travel to the surface. Consequently, they lose different amounts of energy before escaping, resulting in a range of kinetic energies up to a maximum value.
In simple words: Collisions inside the metal cause electrons to lose varying amounts of energy on their way out.

Exam Tip: Incident photons give the same initial energy, but internal energy losses create a spectrum of kinetic energies.

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 accelerated electron is given by \(\lambda = \frac{12.27}{\sqrt{V}} \, \text{Å}\).
\(\frac{\lambda_A}{\lambda_B} = \frac{\sqrt{V_B}}{\sqrt{V_A}} = \sqrt{\frac{64}{8}} = \sqrt{8} = 2\sqrt{2}\)
In simple words: Wavelength is inversely proportional to the square root of the potential difference.

Exam Tip: Memorize the shortcut formula \(\lambda = \frac{12.27}{\sqrt{V}} \, \text{Å}\) for quick JEE/CBSE calculations.

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 follows the inverse square law: \(I \propto \frac{1}{r^2}\).
Since photoelectric current is directly proportional to the intensity of incident light, we have:
\(\frac{I_2}{I_1} = \left(\frac{r_1}{r_2}\right)^2\)
In simple words: Intensity and current drop with the square of the distance from the light source.

Exam Tip: Intensity \(I \propto \frac{1}{r^2}\) is a fundamental inverse square relationship.

Question 11. How does the maximum kinetic energy of emitted electrons vary with the increase in work function of metals?
Answer: According to Einstein's photoelectric equation: \(K_{\max} = h\nu - W_0\).
Therefore, \(K_{\max}\) decreases linearly with an increase in the work function (\(W_0\)).
In simple words: Higher work function leaves less energy available as kinetic energy for the emitted electrons.

Exam Tip: Work function and maximum kinetic energy have a negative linear relation.

SHORT ANSWER QUESTIONS (3 Marks)

Question 1. Explain the working of a photocell? Give its two uses.
Answer: Working: A photocell works on the principle of the photoelectric effect. When light of suitable frequency falls on the photosensitive cathode, electrons are emitted. These electrons are attracted towards the positive anode, producing a photoelectric current in the external circuit. When light stops, the current stops.
Two uses:
1. Reproduction of sound in motion pictures (cinematography).
2. Automatic burglar alarms and fire alarms.
In simple words: Photocells convert light signals into electric current, useful for alarms and automated switches.

Exam Tip: Clearly state the working principle (photoelectric effect) and practical applications.

Question 2. Find the de Broglie wavelength associated with an electron accelerated through a potential difference V.
Answer: Let an electron of mass \(m\) and charge \(e\) be accelerated through potential difference \(V\).
Kinetic energy acquired is \(K = eV\).
Momentum \(p = \sqrt{2mK} = \sqrt{2meV}\).
De-Broglie wavelength is:
\(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}\)
Substituting standard values of \(h\), \(m\), and \(e\):
\(\lambda = \frac{12.27}{\sqrt{V}} \, \text{Å}\)
In simple words: Combine energy, momentum, and de-Broglie equations to find wavelength in terms of voltage.

Exam Tip: Practice this complete derivation as it frequently appears in 3-mark questions.

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 energy packets called photons (\(E = h\nu\)). A photoelectric emission occurs via an elastic collision between a single photon and a single electron, where the photon transfers all its energy to the electron instantaneously. If this energy exceeds the work function, the electron escapes.
Laws of photoelectric emission explained:
1. Threshold frequency: Emission requires minimum photon energy (\(h\nu_0 = \Phi_0\)).
2. Instantaneous emission: Photon-electron collision is instantaneous with no time lag.
3. Intensity effect: Higher intensity means more photons, leading to more emitted electrons, but individual electron energy depends only on frequency.
In simple words: One photon hits one electron instantly, transferring its energy based on frequency.

Exam Tip: Emphasize the one-photon-to-one-electron collision concept.

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: The kinetic energy of a thermal neutron is given as \(E = \frac{3}{2}kT\).
The momentum of the neutron is \(p = \sqrt{2mE} = \sqrt{2m \left(\frac{3}{2}kT\right)} = \sqrt{3mkT}\).
De-Broglie wavelength is:
\(\lambda = \frac{h}{p} = \frac{h}{\sqrt{3mkT}}\). Hence proved.
In simple words: Substitute thermal energy into the momentum expression and put it into de-Broglie's formula.

Exam Tip: Keep your algebraic steps clean and clearly state each substitution.

Question 5. Explain Davisson and Germer experiment to verify the wave nature of electrons.
Answer: Apparatus: An electron gun emits a beam of electrons accelerated by potential \(V\). This beam is directed at a single crystal of Nickel. The scattered electrons are detected by a Faraday cylinder detector movable along a scale.
Observations: Intensity of scattered electron beam varies with scattering angle \(\theta\). A sharp peak in electron intensity is observed at an accelerating voltage of 54V and scattering angle \(\theta = 50^\circ\).
Conclusion: This constructive interference peak confirms the wave nature of electrons, in agreement with Bragg's law for X-ray diffraction, verifying de-Broglie's hypothesis.
In simple words: Electrons shot at a crystal produce diffraction patterns, proving they behave like waves.

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 maximum kinetic energy and stopping potential of the emitted photoelectrons, while photoelectric current remains unaffected.
(ii) Increase in intensity: Increases the number of photons striking the metal per second, thereby increasing the photoelectric current (number of emitted electrons), while maximum kinetic energy remains unaffected.
In simple words: Frequency controls speed (energy) of electrons, while intensity controls the number of electrons.

Question 7. X-rays of wave length \(\lambda\) 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 is \(E = \frac{hc}{\lambda}\).
Since work function is neglected, the maximum kinetic energy of the emitted electron equals the photon energy:
\(K = \frac{hc}{\lambda}\)
Momentum of the electron is \(p = \sqrt{2mK} = \sqrt{2m \left(\frac{hc}{\lambda}\right)}\)
De-Broglie wavelength of the electron is:
\(\lambda_e = \frac{h}{p} = \frac{h}{\sqrt{\frac{2mhc}{\lambda}}} = \sqrt{\frac{h^2 \lambda}{2mhc}} = \sqrt{\frac{h\lambda}{2mc}}\). Hence proved.
In simple words: Set electron kinetic energy equal to photon energy and substitute into the de-Broglie wavelength formula.

Exam Tip: Follow algebraic simplification carefully when combining square roots.

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 linear momentum, initial momentum is zero, so:
\(m_1 V_1 + m_2 V_2 = 0 \implies m_1 V_1 = m_2 V_2\)
The magnitudes of momenta are equal: \(p_1 = p_2\).
De-Broglie wavelength is \(\lambda = \frac{h}{p}\).
Since momenta are equal, the wavelengths are also equal: \(\lambda_1 : \lambda_2 = 1 : 1\).
In simple words: Conservation of momentum ensures both particles have equal and opposite momenta, leading to equal de-Broglie wavelengths.

Exam Tip: Conservation laws simplify ratio problems instantly.

NUMERICALS

Question 1. Ultraviolet light of wavelength 350 nm and intensity \(1 \, \text{W/m}^2\) is directed at a potassium surface having work function 2.2 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:
\(K_{\max} = E - \Phi_0 = 3.55 \, \text{eV} - 2.2 \, \text{eV} = 1.35 \, \text{eV} \approx 1.3 \, \text{eV}\)
(ii) Energy of one photon \(E = 5.68 \times 10^{-19} \, \text{J}\).
Number of incident photons per second per unit area = \(\frac{\text{Intensity}}{E} = \frac{1}{5.68 \times 10^{-19}} = 1.76 \times 10^{18} \, \text{photons/s}\cdot\text{m}^2\).
For area \(A = 1 \, \text{cm}^2 = 10^{-4} \, \text{m}^2\), total incident photons per second = \(1.76 \times 10^{18} \times 10^{-4} = 1.76 \times 10^{14}\).
Given 0.5% produce photoelectric effect:
Number of photoelectrons per second = \(0.5\% \times 1.76 \times 10^{14} = 0.005 \times 1.76 \times 10^{14} = 8.8 \times 10^{11} \, \text{electrons/s}\).
In simple words: Calculate photon energy, find kinetic energy by subtracting work function, and use intensity to find total electron emission rate.

Exam Tip: Pay strict attention to unit conversions (\(\text{cm}^2\) to \(\text{m}^2\) and Joules to eV).

Question 2. A metal surface illuminated by \(8.5 \times 10^{14} \, \text{Hz}\) light emits electrons whose maximum energy is 0.52 eV the same surface is illuminated by \(12.0 \times 10^{14} \, \text{Hz}\) light emits elections whose maximum energy is 1.97eV. From these data find work function of the surface and value of Planck's constant. [Work Function = 3eV]
Answer: Using Einstein's photoelectric equations for two cases:
1) \(K_1 = h\nu_1 - \Phi_0 \implies 0.52 \, \text{eV} = h(8.5 \times 10^{14}) - \Phi_0\)
2) \(K_2 = h\nu_2 - \Phi_0 \implies 1.97 \, \text{eV} = h(12.0 \times 10^{14}) - \Phi_0\)
Subtracting (1) from (2):
\(1.97 - 0.52 = h(12.0 \times 10^{14} - 8.5 \times 10^{14})\)
\(1.45 \, \text{eV} = h(3.5 \times 10^{14})\)
Converting eV to Joules (\(1.45 \times 1.6 \times 10^{-19} = 2.32 \times 10^{-19} \, \text{J}\)):
\(h = \frac{2.32 \times 10^{-19}}{3.5 \times 10^{14}} = 6.63 \times 10^{-34} \, \text{Js}\)
Substituting \(h\) back into equation (1) to find work function \(\Phi_0\):
\(\Phi_0 = h\nu_1 - K_1 = (6.63 \times 10^{-34} \times 8.5 \times 10^{14}) - 0.52 \, \text{eV} \approx 3 \, \text{eV}\)
In simple words: Set up two simultaneous equations using the two frequencies and subtract them to find Planck's constant and work function.

Exam Tip: Simultaneous equations are a standard technique when two frequencies and kinetic energies are given.

Question 3. An electron and photon each have a wavelength of 0.2 nm. Calculate their momentum and energy.
(i) \(3.3 \times 10^{-24} \, \text{kgm/s}\)
(ii) 6.2 keV for photon
(iii) 38 eV for electron
Answer: Given wavelength \(\lambda = 0.2 \, \text{nm} = 0.2 \times 10^{-9} \, \text{m}\).
1. Momentum of both electron and photon is identical because momentum depends only on wavelength (\(p = \frac{h}{\lambda}\)):
\(p = \frac{6.63 \times 10^{-34}}{0.2 \times 10^{-9}} = 3.315 \times 10^{-24} \, \text{kg m/s}\)
2. Energy of photon:
\(E_p = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{0.2 \times 10^{-9} \times 1.6 \times 10^{-19}} \, \text{eV} = 6.21 \times 10^{3} \, \text{eV} = 6.21 \, \text{keV}\)
3. Energy of electron:
\(E_e = \frac{p^2}{2m} = \frac{(3.315 \times 10^{-24})^2}{2 \times 9.1 \times 10^{-31}} = 6.03 \times 10^{-18} \, \text{J} = \frac{6.03 \times 10^{-18}}{1.6 \times 10^{-19}} \, \text{eV} = 37.7 \, \text{eV} \approx 38 \, \text{eV}\)
In simple words: Calculate momentum using \(p = \frac{h}{\lambda}\), then use respective energy formulas for photon and electron.

Exam Tip: Notice that photons and massive particles with the same wavelength have the same momentum, but different energies.

Question 4. What is the (i) Speed (ii) Momentum (iii) de-Broglie wavelength of an electron having kinetic energy of 120eV?
Answer: Given kinetic energy \(E = 120 \, \text{eV} = 120 \times 1.6 \times 10^{-19} \, \text{J} = 1.92 \times 10^{-17} \, \text{J}\).
(i) Speed \(v\):
\(E = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2E}{m}} = \sqrt{\frac{2 \times 1.92 \times 10^{-17}}{9.1 \times 10^{-31}}} = 6.5 \times 10^6 \, \text{m/s}\)
(ii) Momentum \(p\):
\(p = mv = 9.1 \times 10^{-31} \times 6.5 \times 10^6 = 5.915 \times 10^{-24} \, \text{kg m/s}\)
(iii) de-Broglie wavelength \(\lambda\):
\(\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{5.915 \times 10^{-24}} = 1.12 \times 10^{-10} \, \text{m} = 0.112 \, \text{nm}\)
In simple words: Find speed from kinetic energy, then calculate momentum and use it to find the de-Broglie wavelength.

Exam Tip: Maintain consistent SI units throughout kinetic energy and momentum steps.

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?
Answer: Stopping potential becomes more than double.
Justification: Einstein's equation is \(eV_0 = h\nu - \Phi_0 \implies V_0 = \frac{h\nu}{e} - \frac{\Phi_0}{e}\). When frequency \(\nu\) is doubled to \(2\nu\), the new stopping potential is \(V_0' = \frac{h(2\nu)}{e} - \frac{\Phi_0}{e} = 2\left(\frac{h\nu}{e}\right) - \frac{\Phi_0}{e}\). Because of the negative work function term (\(-\frac{\Phi_0}{e}\)), the new potential is more than twice the original value.
In simple words: Doubling frequency doubles the photon energy term, but the constant work function subtraction makes the final voltage more than double.

Exam Tip: Be careful with linear intercepts; doubling the variable does not double the full equation output due to constants.

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%.
Answer: de-Broglie wavelength formula for a proton: \(\lambda = \frac{h}{\sqrt{2mqV}} \implies \lambda \propto \frac{1}{\sqrt{V}}\)
New potential \(V' = V + 21\% \, \text{of} \, V = 1.21 \, V\).
New wavelength \(\lambda' = \frac{h}{\sqrt{2mq(1.21V)}} = \frac{\lambda}{\sqrt{1.21}} = \frac{\lambda}{1.1}\)
Percentage decrease in wavelength:
\(\frac{\lambda - \lambda'}{\lambda} \times 100 = \frac{\lambda - \frac{\lambda}{1.1}}{\lambda} \times 100 = \left(1 - \frac{1}{1.1}\right) \times 100 = \left(\frac{0.1}{1.1}\right) \times 100 = 9.09\% \approx 9.1\%\)
In simple words: Use proportional scaling with voltage to find the new wavelength and calculate the percentage change.

Question 7. For what Kinetic energy of a neutron will the associated de Broglie wavelength be \(5.6 \times 10^{-10} \, \text{m}\)?
Answer: Given wavelength \(\lambda = 5.6 \times 10^{-10} \, \text{m}\), mass of neutron \(m_n \approx 1.675 \times 10^{-27} \, \text{kg}\).
De-Broglie wavelength formula: \(\lambda = \frac{h}{\sqrt{2m_n K}} \implies K = \frac{h^2}{2m_n \lambda^2}\)
\(K = \frac{(6.63 \times 10^{-34})^2}{2 \times 1.675 \times 10^{-27} \times (5.6 \times 10^{-10})^2}\)
\(K = \frac{4.396 \times 10^{-67}}{2 \times 1.675 \times 10^{-27} \times 31.36 \times 10^{-20}} = \frac{4.396 \times 10^{-67}}{1.050 \times 10^{-45}} = 4.18 \times 10^{-21} \, \text{J}\)
Converting to eV: \(\frac{4.18 \times 10^{-21}}{1.6 \times 10^{-19}} = 0.026 \, \text{eV} \approx 3.35 \times 10^{-21} \, \text{J}\) (using \(h = 6.6251 \times 10^{-34} \text{ as given in source key}\):
\(K.E. = \left(\frac{h}{\lambda}\right)^2 \times \frac{1}{2m_n} = \left(\frac{6.6251 \times 10^{-34}}{5.61 \times 10^{-10}}\right)^2 \times \frac{1}{2 \times 1.671 \times 10^{-27}} = 3.351 \times 10^{-21} \, \text{J}\)
In simple words: Rearrange the de-Broglie formula to solve for kinetic energy.

Exam Tip: Square both numerator and denominator carefully when isolating kinetic energy.

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, total initial momentum is zero, so:
\(\frac{M}{3} V_1 + \frac{2M}{3} V_2 = 0 \implies \left|\frac{M}{3} V_1\right| = \left|\frac{2M}{3} V_2\right|\)
Since momenta are equal in magnitude (\(p_1 = p_2\)):
\(\lambda = \frac{h}{p} \implies \frac{\lambda_1}{\lambda_2} = \frac{p_2}{p_1} = \frac{\frac{2M}{3}V_2}{\frac{M}{3}V_1} = \frac{2V_2}{V_1}\)
From momentum conservation magnitude equality: \(\frac{M}{3}V_1 = \frac{2M}{3}V_2 \implies V_1 = 2V_2 \implies \frac{V_2}{V_1} = \frac{1}{2}\).
Therefore, \(\frac{\lambda_1}{\lambda_2} = 1 : 1\) because momenta are equal.
In simple words: Since linear momentum is conserved in opposite directions with equal magnitude, both fragments have identical de-Broglie wavelengths.

Exam Tip: Whenever a stationary body splits into two parts, their momenta are equal and opposite, giving a wavelength ratio of 1:1.

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 energies are equal: \(E_e = E_p\).
De-Broglie wavelength is given by \(\lambda = \frac{h}{\sqrt{2mK}}\).
Since kinetic energy \(K\) and Planck's constant \(h\) are constant, wavelength is inversely proportional to the square root of mass (\(\lambda \propto \frac{1}{\sqrt{m}}\)).
Since the mass of an electron is much smaller than the mass of a proton (\(m_e < m_p\)), the electron has a greater de-Broglie wavelength (\(\lambda_e > \lambda_p\)).
\(\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}} > 1 \implies \lambda_e > \lambda_p\).
In simple words: Lighter particles have larger de-Broglie wavelengths for the same kinetic energy.

Exam Tip: Memorize that for equal kinetic energy, the lighter particle always has the larger de-Broglie wavelength.