If the frequency of incident radiation on a photocell is doubled for t

1.	If the frequency of incident radiation on a photocell is doubled for the same intensity, what changes will you observe in :- i. Kinetic energy of photoelectron emitted, ii. Photoelectric current and iii. Stopping potential. Justify your answer in each case.
Answer» (i). When the frequency of the incident light is doubled, the kinetic energy of emitted photoelectron becomes more than double. Let E₁ and E₂ be the K.E. of photoelectrons for incident light of frequency ν and 2ν respectively. Then, hν = E₁+ ϕ₀ and 2hν = E₂+ ϕ₀ ∴ 2 = \(\frac{E_2+ϕ_0}{E_2 +ϕ_0}\) Or, 2E₁ + 2ϕ₀ = E₂ + ϕ₀ Or, E₂ + 2E₁+ ϕ₀ Or, E₂ > 2E₁ (ii). Photoelectric current remains unchanged as it depends upon the intensity of incident light. (iii). The stopping potential becomes more than double. eV₀ = (K.E.)_max = hν - ϕ₀ or, hν = e₀ + ϕ₀ and eV₀′ = 2hν - ϕ₀ = 2(eV₀ + ϕ₀) − ϕ₀ = 2 eV₀ + ϕ₀ ∴V₀′> 2₀

If the frequency of incident radiation on a photocell is doubled for the same intensity, what changes will you observe in :- i. Kinetic energy of photoelectron emitted, ii. Photoelectric current and iii. Stopping potential. Justify your answer in each case.

Answer»

(i). When the frequency of the incident light is doubled, the kinetic energy of emitted photoelectron becomes more than double.

Let E₁ and E₂ be the K.E. of photoelectrons for incident light of frequency ν and 2ν respectively.

Then,

hν = E₁+ ϕ₀ and

2hν = E₂+ ϕ₀

∴ 2 = \(\frac{E_2+ϕ_0}{E_2 +ϕ_0}\)

Or, 2E₁ + 2ϕ₀ = E₂ + ϕ₀

Or, E₂ + 2E₁+ ϕ₀

Or, E₂ > 2E₁

(ii). Photoelectric current remains unchanged as it depends upon the intensity of incident light.

(iii). The stopping potential becomes more than double.

eV₀ = (K.E.)_max = hν - ϕ₀

or, hν = e₀ + ϕ₀ and

eV₀′ = 2hν - ϕ₀

= 2(eV₀ + ϕ₀) − ϕ₀

= 2 eV₀ + ϕ₀

∴V₀′> 2₀