In a photoelectric experiment ultraviolet light of wavelength 280 nm is used with lithium cathode having work function \(\phi=2.5 \mathrm{eV}\). If the wavelength of incident light is switched to 400 nm , find out the change in the stopping potential. \(\left(\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}, \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}\right)\)

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Modern Physics

In a photoelectric experiment ultraviolet light of wavelength 280 nm is used with lithium cathode having work function

ϕ = 2.5 eV

. If the wavelength of incident light is switched to 400 nm , find out the change in the stopping potential.

(h = 6.63 \times 1 0^{- 34} Js, c = 3 \times 1 0^{8} ms^{- 1})

Answer: (A)

The stopping potential obeys

e V_{s} = \frac{h c}{λ} - ϕ

, and since the cathode is unchanged the work function cancels in the difference:

e Δ V_{s} = h c (\frac{1}{λ _{1}} - \frac{1}{λ _{2}})

With

λ_{1} = 280

nm and

λ_{2} = 400

nm,

\frac{h c}{e} = \frac{6.63 \times 1 0 ^{- 34} \times 3 \times 1 0 ^{8}}{1.6 \times 1 0 ^{- 19}} = 1.243 \times 1 0^{- 6} V m

Δ V_{s} = 1.243 \times 1 0^{- 6} (\frac{1}{280 \times 1 0 ^{- 9}} - \frac{1}{400 \times 1 0 ^{- 9}}) = 1.243 \times 1 0^{- 6} \times 1.071 \times 1 0^{6} \approx 1.3 V

Light of

400

nm carries

3.1

eV, still above the

2.5

eV work function, so emission continues and the stopping potential falls by

1.3

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Single correct option