When light of wavelength ' λ ' is incident on a photosensitive surface, the stopping potential is ' V '. When light of wavelength ' 2λ ' is incident on the same surface, the stopping potential is V/3 . Threshold

Question

Quantrex Academy · Accepted Answer

Correct answer: C. 4λ

Using Einstein's photoelectric equation:

eV = \dfrac{hc}{\lambda} - \phi

\dfrac{eV}{3} = \dfrac{hc}{2\lambda} - \phi

Multiplying the second equation by

3

:

eV = \dfrac{3hc}{2\lambda} - 3\phi

Equating the two expressions for

eV

:

\dfrac{hc}{\lambda} - \phi = \dfrac{3hc}{2\lambda} - 3\phi

2\phi = \dfrac{3hc}{2\lambda} - \dfrac{hc}{\lambda}

2\phi = \dfrac{hc}{2\lambda}

\phi = \dfrac{hc}{4\lambda}

The work function is given by

\phi = \dfrac{hc}{\lambda_0}

, where

\lambda_0

is the threshold wavelength.

\dfrac{hc}{\lambda_0} = \dfrac{hc}{4\lambda}

\lambda_0 = 4\lambda

Answer:

4\lambda

When light of wavelength ' λ ' is incident on a photosensitive...

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