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Q.

The stopping potential for electrons from a photosensitive surface illuminated by light of wavelength 491 nm is 0.710V. When the incident wavelength is changed to a new value, the stopping potential is 1.43V. The new wavelength is:

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a

309 nm

b

329 nm

c

382 nm

d

400 nm

answer is C.

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Detailed Solution

$\begin{array}{l} Energy of photon, E = \frac{1240}{λ (n m)} eV \\ \Rightarrow E = \frac{1240}{491} eV \\ \Rightarrow E = 2.53 eV \\ Work function = ϕ_{0} = E - V_{0} e = 2.53 - 0.710 = 1.82 eV \end{array}$

$\begin{array}{l} F o r c a s e 2, w o r k f u n c t i o n w i l r e m a i n s a m e \\ \Rightarrow E = \frac{h c}{λ} - ϕ o \\ \Rightarrow V_{o} e = E - ϕ_{o} \\ \Rightarrow 1.43 + ϕ_{o} = E \\ \Rightarrow 1.43 + 1.82 = E \\ \Rightarrow E = 3.25 e V \\ S o, λ = \frac{1240}{3.25} n m = 382 n m \end{array}$

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The stopping potential for electrons from a photosensitive surface illuminated by light of wavelength 491 nm is 0.710V. When the incident wavelength is changed to a new value, the stopping potential is 1.43V. The new wavelength is: