JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Modern Physics

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A monochromatic neon lamp with wavelength of 670.5 nm illuminates a photo-sensitive material which has a stopping voltage of 0.48 V. What will be the stopping voltage if the source light is changed with another source of wavelength of 474.6 nm ?

(A)1.5 V
(B)0.24 V
(C)0.96 V
(D)1.25 V✓ Correct

Answer: (D)

Solution

Using Einstein's photoelectric equation:

e V_{s} = h ν - ϕ = \frac{h c}{λ} - ϕ

. For the first wavelength:

e (0.48) = \frac{h c}{670.5} - ϕ

. For the second wavelength:

e V_{s}^{'} = \frac{h c}{474.6} - ϕ

. Subtracting:

e (V_{s}^{'} - 0.48) = h c (\frac{1}{474.6} - \frac{1}{670.5})

. Using

h c = 1240

eV·nm:

V_{s}^{'} - 0.48 = 1240 (\frac{1}{474.6} - \frac{1}{670.5}) = 1240 (0.002108 - 0.001492) = 1240 (0.000616) \approx 0.764

V. Therefore,

V_{s}^{'} \approx 1.24

\approx 1.25

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Modern Physics

Solution

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