## Dual Nature of Radiation and Matter

### Maharashtra Board-Class-12th-Physics-Chapter-14

### Solutions

**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 photo cell is**

**(A) zinc **

**(B) aluminum**

**(C) nickel **

**(D) potassium**

(D) potassium

Explanation:

- A photocell is a sensor that changes its resistance based on light intensity.
- It consists of cathodes and anodes, with the cathode emitting electrons and the anode collecting them.
- The element that is used as a cathode should have less ionization energy so that it can emit an electron when subjected to light. Potassium has lower ionization energy. Hence it is used in photocells as the cathode.

**(ii) Polychromatic (containing many different frequencies) radiation is used in an experiment on 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**

(C) will depend on the shortest wavelength

Explanation:

- The photoelectric effect is the emission of electrons when a photon hits metallic surfaces, resulting in the emission of photoelectrons.
- The rate of photoelectrons emission depends on the frequency or wavelength of the incident light.
- Higher frequency or lower wavelength light causes more photoelectrons, resulting in photocurrent.
- The stopping potential is the minimum required to prevent photoelectrons from ejecting due to low wavelength radiation.
- Polychromatic light with high frequency and low wavelength contributes to electron ejection.

**(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**

(A) electron

**(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} **

**≈**

*N*

_{Blue}(C) *N*_{Red} > *N*_{Blue}

Explanation:

The energy of photons emitted in a unit of time is equal to the power and dimensions of the photons. Blue light, due to its higher frequency, emits more energy than red light, resulting in a smaller required number of blue photons.

**(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**

**(C) is valid for a photon but not for an electron**

Explanation:

The equation E = pc, an extension of Einstein's mass−energy relation, is valid when the particle's rest mass is zero. However, the electron's rest mass is not zero, making E = pc not valid for the electron but valid for the photon.

**Question 2. Answer in brief.**

**(i) What is photoelectric effect?**

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.

**(ii) Can microwaves be used in the experiment on photoelectric effect?**

No

**Reason **: The microwave frequency is in the range of 10^{9} Hz to 10^{12} Hz. This frequency range is insufficient to provide energy for the photoelectric effect.

**(iii) Is it always possible to see photoelectric effect with red light?**

No,

**Reason :**

Red light, which behaves like a wave, can overcome an electron's work function with high intensity or for a long time due to its particle behavior. The energy required to remove an electron from its orbital equals the electron's work function, which is not available in red light due to its particle nature.

**(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.**

**Table 14.1 : Typical values of work function for some common metals.**

Metal |
Work function (in eV) |

Potassium |
2.3 |

Sodium |
2.4 |

Calcium |
2.9 |

Zinc |
3.6 |

Silver |
4.3 |

Aluminum |
4.3 |

Tungsten |
4.5 |

Copper |
4.7 |

Nickel |
5 |

Gold |
5.1 |

Based on the metal's work function values in the table above, we can conclude that gold will require the highest frequency of incident radiation to generate photocurrent.

**(v) What do you understand by the term wave−particle duality? Where does it apply?**

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 / Schrödinger 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.

**Question 3. Explain the inverse linear dependence of stopping potential on the incident wavelength in a photoelectric effect experiment.**

We have eV_{0} = \(\frac{hc}{λ}\) − ø 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 1/λ increases, V_{0} increases.

The plot of V_{0} verses 1/λ 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.

**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.**

- Electromagnetic radiation with a frequency greater than the threshold frequency on a metal surface results in electron emission.
- However, not every photon is effective in liberating electrons, and the number of electrons emitted per second is significantly lower than the number of photons incident per second.
- Uneffective photons are reflected, scattered, or absorbed, causing a rise in the metal surface temperature.
- The maximum kinetic energy of a photoelectron depends on the incident radiation frequency and threshold frequency, not the intensity of the radiation.
- The increase in intensity results in increase in the number of electrons emitted per second.

**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.**

Under certain conditions an electron exhibits wave nature. Waves associated with a moving electron are called matter waves or de Broglie waves or Schrödinger waves. The de Broglie wavelength of these matter waves is given by λ = 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.

**Question 6. State the importance of Davisson and Germer experiment**.

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.

**Question 7. What will be the energy of each photon in monochromatic light of frequency 5 × 10 ^{14} Hz?**

**Given** : v = 5 x 10^{14} Hz, h = 6.63 x 10^{−34} J.s, 1eV = 1.6 x 10^{−19} J

The energy of each photon,

E = hv = (6.63 x 10^{−34})(5 x 10^{14}) = 3.315 × 10^{−19} J = \(\frac{3.315×10^{–19}}{1.6×10^{–19}}\) eV = **2.072 eV **

**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 4.1×10 ^{−15} V s. Given that the charge of an electron is 1.6 × 10^{−19} C, find the value of the Planck’s constant h.**

**Given** : Slope = 4.1 x 10^{–15} V.s, e = 1.6 × 10^{−}^{19} C

V_{0}e = hν – hν_{0}

∴ V_{0} = \((\frac{h}{e})\)ν – \((\frac{hν_0}{e})\)

∴ Slope = h/e

∴ Planck's constant, h = Slope x e = (4.1 x 10^{–15})(1.6 × 10^{−}^{19}) = **6.56 x 10 ^{–34} J.s **…(as 1V = )

**Question 9. The threshold wavelength of tungsten is 2.76 × 10 ^{−5} cm. **

**(a) Explain why no photoelectrons are emitted when the wavelength is more than 2.76 × 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 × 10 ^{−5} cm and**

**(ii) radiation of frequency 4 × 10 ^{15} Hz is made incident on the tungsten surface.**

**Given** : 10 = 2.76 x 10^{–5} cm = 2.76 x 10^{–7} m, λ = 1.80 × 10^{–5}cm = 1.80 × 10^{–7} m, ν = 4 x 10^{15} Hz, h = 6.63 x 10^{–34} J.s, c = 3 × 10^{8} m/s

(a) For λ > λ_{0}, ν < ν_{0} (threshold frequency).

∴ hν < hν_{0}. Hence, no photoelectrons are emitted.

(b) Maximum kinetic energy of electrons ejected = hc\((\frac{1}{λ}–\frac{1}{λ_0})\)

= (6.63 x 10^{–34})(3 × 10^{8})\((\frac{10^7}{1.8}–\frac{10^7}{2.76})\)

= (6.63 x 10^{–19})(0.5555–0.3623) = **1.281 x 10 ^{–19} J**

= \(\frac{1.281×10^{–19}}{1.6×10^{–19}}\) eV = **0.8006 eV**

(c) Maximum kinetic energy of electrons ejected = hν – \(\frac{hc}{λ_0}\)

= (6.63 x 10^{–34})(4 × 10^{15}) − \(\frac{(6.63×10^{–34})(3×10^8)}{2.76×10^{–7}}\)

= 26.52 x 10^{–19} – 7.207 x 10^{–19}

=** 19.313 ****x 10 ^{–19} J**

= \(\frac{19.313×10^{–19}}{1.6×10^{–19}}\) eV = **12.07 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.**

**Given** : V_{0} = 0.8 V, λ = 4950 Å = 4.950 x 10^{−}^{7} m, V_{0}' = 1.2 V, h = 6.63 x 10^{−}^{34} J.s, c = 3 x 10^{8} m/s.

V_{0}e = hν – ø = \(\frac{hc}{λ}\) – ø

∴ The work function of the cathode material,

ø = \(\frac{hc}{λ}\) – V_{0}e

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

= 4.018 x 10^{−}^{19}– 1.28 × 10^{−}^{19} = 2.738 × 10^{−}^{19 }J = \(\frac{2.738×10^{–19}}{1.6×10^{–19}}\) eV = **1.711 eV**

(ii) V_{0}’e = \(\frac{hc}{λ'}\) – ø

∴ \(\frac{hc}{λ'}\) = V_{0}’e + ø

∴ The wavelength of the second source,

λ' = \(\frac{hc}{V_0'e+ø}\)

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

= \(\frac{19.89×10^{–26}}{4.658×10^{–19})}\)

= 4.2750 × 10** ^{−7} **m

**= 4270**

**Å**

**Question 11. Radiation of wavelength 4500 Å is incident on a metal having work function 2.0 eV. Due to the presence of a magnetic field B, the most energetic photoelectrons 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?**

**Given** : λ = 4500 Å = 4.5 x 10^{–7} m, ø = 2.0 eV = 2 × 1.6 × 10^{–19} J = 3.2 × 10^{–19} J, h = 6.63 x 10^{–34} J.s, c = 3 x 10^{8} m/s, r = 20 cm = 0.2m, e = 1.6 ×10^{–19} C,

m = 9.1 × 10^{–31} kg

mv^{2}_{max} = KE_{max} = \(\frac{hc}{λ}\) – ø = \(\frac{(6.63×10^{–34})(3×10^8)}{4.5×10^{–7}}\) – 3.2 × 10^{–19}

= 4.42 10^{–19} – 3.2 × 10^{–19 }= 1.22 × 10^{–19} J

∴ v_{max} = \(\sqrt{\frac{2}{m}KE_{max}}\) = \(\sqrt{\frac{2×1.22×10^{–19}}{9.1×10^{–31}}}\)

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

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

∴ B = \(\frac{mv_{max}}{re}\) = \(\frac{(9.1×10^{–31})(5.178×10^5)}{(0.2)(1.6×10^{–19})}\) **= 1.472 ****x 10 ^{–5} T**

This is the value of the magnetic field.

**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?**

Incident wavelength (in Å) |
2536 |
3650 |

Stopping potential (in V) |
1.95 |
0.5 |

**Given** : λ = 2536 Å = 2.536 x 10^{–7} m, λ' = 3650 A = 3.650 x 10^{–7} m, V_{0} = 1.95 V, V_{0}' = 0.5 V, c = 3 × 10^{8} m/s, e = 1.6 × 10^{–19} C

**(i) **V_{0}e = \(\frac{hc}{λ}\) – ø and V_{0}’e = \(\frac{hc}{λ'}\) – ø

∴ (V_{0 }– V_{0}’)e = hc\((\frac{1}{λ}–\frac{1}{λ'})\)

∴ (1.95 – 0.5)( 1.6 × 10^{–19}) = h(3 × 10^{8})\((\frac{10^7}{2.536}–\frac{10^7}{3.650})\)

∴ 2.32 x 10^{–19 }= h(3 × 10^{8})(0.3943 – 0.2740)

∴ h = \(\frac{2.32×10^{–34}}{0.3609}\) = **6.428 ****x ****10 ^{–34} J.s**

This is the value of Planck's constant.

(ii) ø = \(\frac{hc}{λ}\) – V_{0}e

= \(\frac{(6.428×10^{–34})(3×10^8)}{2.536×10^{–7}}\) – (1.95)(1.6 × 10^{−}^{19})

= 7.604 x 10^{−}^{19}– 3.12 × 10^{−}^{19} = 4.484 × 10^{−}^{19 }J

= \(\frac{4.484×10^{–19}}{1.6×10^{–19}}\) eV = **2.803 eV**

This is the work function of the cathode material.

(iii) ø = hν_{0}

∴ The threshold frequency, ν_{0 }= \(\frac{ø}{h}\) = \(\frac{4.484×10^{–19}}{6.428×10^{–34}}\) = **6.976 ****× 10 ^{14 }Hz**

(iv) ν_{0} = \(\frac{c}{λ_0}\),

∴ The threshold wavelength, λ_{0 }= \(\frac{c}{ν_0}\) = \(\frac{3×10^8}{6.976×10^{14}}\) = 4.3 × 10^{–7 }m = 4300 Å

(v) The most likely metal used for emitter : calcium.

**Question 13. Calculate the wavelength associated with an electron, its momentum and speed**

**(a) when it is accelerated through a potential of 54 V,**

**(b) when it is moving with kinetic energy of 150 eV.**

**Given** : V = 54 V, m = 9.1 x 10^{–31} kg, e = 1.6 x 10^{–19} C, h = 6.63 x 10^{–34} J.s,

KE = 150 eV

(a) We assume that the electron is initially at rest.

∴ V_{e }= \(\frac{1}{2}\)mv^{2}

∴ v = \(\sqrt{\frac{2ve}{m}}\) = \(\sqrt{\frac{2(54)(1.6×10^{–19}}{9.1×10^{–31}}}\) = \(\sqrt{19×10^{12}}\) = 4.359 x 10^{6 }m/s

This is the speed of the electron.

p = mv = (9.1 × 10^{–31})(4.359 × 10^{6}) = 3.967 × 10^{–24} kg·m/s

This is the momentum of the electron. The wavelength associated with the electron,

λ = \(\frac{h}{p}\) = \(\frac{6.63×10^{–34}}{3.967×10^{–24}}\) = 1.671 x 10^{–10 }m = **1.671 ****Å = 0.1671 nm**

(b) As KE ∝ \(\sqrt{V}\), we get

\(\frac{v'}{v}\) = \(\sqrt{\frac{15}{54}}\) = 1.666

∴ v’ = 1.666 v = (1.666) x (4.359 x 10^{6}) = 7.262 x 10^{6 }m/s

This is the speed of the electron.

p' = mv' = (9.1 × 10^{–31})(7.262 × 10^{6}) = 6.608 × 10^{–24} kg.m/s

This is the momentum of the electron. The wavelength associated with the electron,

λ' = (\frac{h}{p'}\) = \(\frac{6.63×10^{–34}}{6.608×10^{–24}}\) = 1.003 x 10^{–10} m =**1.003 Å = 0.1003 nm**

**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?**

**Given** : λ (electron)= λ (proton), m (proton)= 1836 m(electron)

(i) λ = \(\frac{h}{p}\), As λ (electron)= λ (proton)

\(\frac{p(electron)}{p(proton)}\) = 1 where p denotes the magnitude of momentum.

(ii) Assuming v << c,

KE = \(\frac{1}{2}\)mv^{2} = \(\frac{m^2v^2}{2m}\) = \(\frac{p^2}{2m}\)

∴ \(\frac{KE(electron)}{KE(proton)}\) = \(\frac{m(proton)}{m(electron)}\) = 1836 as p is the same for the electron and the proton.

**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?**

λ_{1} = λ_{2} v_{1} = 4v_{2}

λ = \(\frac{h}{p}\) = \(\frac{h}{mv}\), ∴ λ_{1} = \(\frac{h}{m_1v_1}\), λ_{2} = \(\frac{h}{m_2v_2}\)

as λ_{1} = λ_{2}, m_{1}v_{1} = m_{2}v_{2}

∴ m_{1} = m_{2} \(\frac{v_2}{v_1}\) = m_{2}\((\frac{1}{4})\) = \(\frac{m_2}{4}\)

As particle 2 is the a–particle, particle 1 (having the mass \(\frac{1}{4}\) times that of the ∝–particle) may be a proton or neutron.

**Question 16. What is the speed of a proton having de Broglie wavelength of 0.08 Å?**

**Given** : λ = 0.08 Å = 8 x 10^{–12} m, h = 6.63 x 10^{–34} J.s, m = 1.672 × 10^{–27} kg

λ = \(\frac{h}{mv}\), ∴ v = \(\frac{h}{λm}\) = \(\frac{6.63×10^{–34}}{(8×10^{–12} )(1.672×10^{–27})}\) = **4.957 x 10 ^{4 }m/s**

This is the speed of the proton.

**Question 17. In nuclear reactors, neutrons travel with energies of 5 × 10 ^{−21} J. Find their speed and wavelength.**

**Given** : KE = 5 x 10^{–21} J, m = 1.675 × 10^{–27} kg, h = 6.63 x 10^{–34} J.s

KE = \(\frac{1}{2}\)mv^{2} = 5 x 10^{–21} J

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

This is the speed of the neutrons. The de Broglie wavelength associated with the neutron,

λ = \(\frac{h}{mv}\) = \(\frac{6.63×10^{–34}}{(1.675×10^{–27})(2.443×10^3)}\) **= 1.620 ****x 10 ^{–10 }m = 1.620 Å**

**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?**

**Given** : m_{p} = 1836 m_{e}

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

\(\frac{λ_e}{λ_p}\) = \((\frac{m_p}{m_e})=(\frac{v_p}{v_e})\) = 1836 as v_{p} = v_{e}.

Thus λ_{ p} < λ_{ e}

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

∴ \(\frac{λ_e}{λ_p}\) = \(\sqrt{\frac{m_pK_p}{m_eK_e}}\) = \(\sqrt{\frac{m_p}{m_e}}\) = \(\sqrt{1836}\) **= 42.85 **…(as K_{p} = K_{e })

Thus λ_{ p} < λ_{ e}

(c) λ = \(\frac{h}{p}\), ∴ \(\frac{λ_e}{λ_p}\) = \(\frac{p_p}{p_e}\) = 1 …(as p_{p} = p_{e})

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