Solutions-Class-12-Physics-Chapter-14-Dual Nature of Radiation and Matter -MSBSHSE

(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.

(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.

(A) electron

(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.

(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.

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.

No

Reason : The microwave frequency is in the range of 10⁹ Hz to 10¹² Hz. This frequency range is insufficient to provide energy for the photoelectric effect.

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.

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.

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.

We have eV₀ = \(\frac{hc}{λ}\) − ø where V₀ 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, V₀ increases.

The plot of V₀ 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.

• 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.

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.

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.

Given : v = 5 x 10¹⁴ 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 10¹⁴) = 3.315 × 10⁻¹⁹ J = \(\frac{3.315×10^{–19}}{1.6×10^{–19}}\) eV = 2.072 eV 

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

V₀e = hν – hν₀

∴ V₀ = \((\frac{h}{e})\)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 = )

Given : 10 = 2.76 x 10^–5 cm = 2.76 x 10^–7 m, λ = 1.80 × 10^–5cm = 1.80 × 10^–7 m, ν = 4 x 10¹⁵ Hz, h = 6.63 x 10^–34 J.s, c = 3 × 10⁸ m/s

(a) For λ > λ₀, ν < ν₀ (threshold frequency).

∴ hν < hν₀. 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 × 10⁸)\((\frac{10^7}{1.8}–\frac{10^7}{2.76})\)

= (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 × 10¹⁵) − \(\frac{(6.63×10^{–34})(3×10^8)}{2.76×10^{–7}}\)

= 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

Given : V₀ = 0.8 V, λ = 4950 Å = 4.950 x 10⁻⁷ m, V₀' = 1.2 V, h = 6.63 x 10⁻³⁴ J.s, c = 3 x 10⁸ m/s.

V₀e = hν – ø = \(\frac{hc}{λ}\) – ø

∴ The work function of the cathode material,

ø = \(\frac{hc}{λ}\) – V₀e

= \(\frac{(6.63×10^{–34})(3×10^8)}{4.950×10^{–7}}\) – (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) V₀’e = \(\frac{hc}{λ'}\) – ø

∴ \(\frac{hc}{λ'}\) = V₀’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

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 10⁸ m/s, r = 20 cm = 0.2m, e = 1.6 ×10^–19 C,

m = 9.1 × 10^–31 kg

mv²_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 10⁵ 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.

Given : λ = 2536 Å = 2.536 x 10^–7 m, λ' = 3650 A = 3.650 x 10^–7 m, V₀ = 1.95 V, V₀' = 0.5 V, c = 3 × 10⁸ m/s, e = 1.6 × 10^–19 C

(i) V₀e = \(\frac{hc}{λ}\) – ø and V₀’e = \(\frac{hc}{λ'}\) – ø

∴ (V₀ – V₀’)e = hc\((\frac{1}{λ}–\frac{1}{λ'})\)

∴ (1.95 – 0.5)( 1.6 × 10^–19) = h(3 × 10⁸)\((\frac{10^7}{2.536}–\frac{10^7}{3.650})\)

∴ 2.32 x 10^–19 = h(3 × 10⁸)(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 = \(\frac{hc}{λ}\) – V₀e

= \(\frac{(6.428×10^{–34})(3×10^8)}{2.536×10^{–7}}\)  – (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 = hν₀

∴ The threshold frequency, ν₀ = \(\frac{ø}{h}\) = \(\frac{4.484×10^{–19}}{6.428×10^{–34}}\) = 6.976 × 10¹⁴ Hz

(iv) ν₀ = \(\frac{c}{λ_0}\),

∴ The threshold wavelength, λ₀ = \(\frac{c}{ν_0}\) = \(\frac{3×10^8}{6.976×10^{14}}\) = 4.3 × 10^–7 m = 4300 Å

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

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 = \(\frac{1}{2}\)mv²

∴ v = \(\sqrt{\frac{2ve}{m}}\) = \(\sqrt{\frac{2(54)(1.6×10^{–19}}{9.1×10^{–31}}}\) = \(\sqrt{19×10^{12}}\) = 4.359 x 10⁶ 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 = \(\frac{h}{p}\) = \(\frac{6.63×10^{–34}}{3.967×10^{–24}}\) = 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 10⁶) = 7.262 x 10⁶ 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'}\) = \(\frac{6.63×10^{–34}}{6.608×10^{–24}}\) = 1.003 x 10^–10 m =1.003 Å = 0.1003 nm

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 = \(\frac{1}{2}\)mv² = \(\frac{m^2v^2}{2m}\) = \(\frac{p^2}{2m}\)

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

λ₁ = λ₂  v₁ = 4v₂

λ = \(\frac{h}{p}\) = \(\frac{h}{mv}\), ∴ λ₁ = \(\frac{h}{m_1v_1}\), λ₂ = \(\frac{h}{m_2v_2}\)

as λ₁ = λ₂,  m₁v₁ = m₂v₂

∴ m₁ = m₂ \(\frac{v_2}{v_1}\) = m₂\((\frac{1}{4})\) = \(\frac{m_2}{4}\)

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.

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 10⁴ m/s

This is the speed of the proton.

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

KE = \(\frac{1}{2}\)mv² = 5 x 10^–21 J

v = \(\sqrt{\frac{2KE}{m}}\) = \(\sqrt{\frac{2(5×10^{–21}}{1.675×10^{–27}}}\) = 2.443 x 10³ 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

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 (\(\frac{1}{2}\)mv²)

∴ \(\frac{λ_e}{λ_p}\) = \(\sqrt{\frac{m_pK_p}{m_eK_e}}\) = \(\sqrt{\frac{m_p}{m_e}}\) = \(\sqrt{1836}\) = 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)