For a certain metal, when monochromatic light of wavelength λ is incident, the stopping potential for photoelectrons is 3V_0 . When the same metal is illuminated by light of wavelength 2λ , then the stopping potential

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Quantrex Academy · Accepted Answer

Correct answer: B. 4

Using Einstein's photoelectric equation:

eV_s = \dfrac{hc}{\lambda} - \Phi

For the first case:

3eV_0 = \dfrac{hc}{\lambda} - \Phi

For the second case:

eV_0 = \dfrac{hc}{2\lambda} - \Phi

Multiplying the second equation by

3

and equating with the first equation:

\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}

Since the work function

\Phi = \dfrac{hc}{\lambda_0}

, where

\lambda_0

is the threshold wavelength:

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

\lambda_0 = 4\lambda

Given

\lambda_0 = \alpha\lambda

, we get

\alpha = 4

.

Answer:

4

For a certain metal, when monochromatic light of wavelength λ is...

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