When a metallic surface is illuminated with radiation of wavelength \lambda, the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2\lambda, the stopping potential is\frac{\text{V}} {\text{4}}. The threshold wavelength for the metallic surface is:

  • Option 1)

    4\lambda

  • Option 2)

    5\lambda

  • Option 3)

    \frac{\text{5}} {2}\lambda

  • Option 4)

    3\lambda

 

Answers (1)

 

Conservation of energy -

h\nu = \phi _{0}+\frac{1}{2}mv^{2}_{max}

h\nu = h\nu _{0}+\frac{1}{2}mv^{2}_{max}

 

h\left ( \nu-\nu _{0} \right )=\frac{1}{2}mv^{2}_{max}

where, h - Plank's constant\ \nu - Frequency\ \nu_{0} - threshold\ frequency\ \phi_{0}- work function

- wherein

 

Let threshold wavelength is {\lambda_{0} }

Then  ev=\frac{hc}{\lambda }-\frac{hc}{\lambda_{0} } ---------------(1)

\frac{ev}{4}=\frac{hc}{2\lambda }-\frac{hc}{\lambda_{0} } -------------------(2)\times 4

\Rightarrow ev=\frac{2hc}{\lambda }-\frac{4hc}{\lambda_{0} } ---------------(3)

From equation (1) and (2)

\frac{2hc}{\lambda }-\frac{4hc}{\lambda_{0} }=\frac{hc}{\lambda }-\frac{hc}{\lambda_{0} }

or

\frac{hc}{\lambda_{0} }=\frac{3hc}{\lambda_{0} } \: or \: \lambda _{0}=3\lambda


Option 1)

4\lambda

This is incorrect option

Option 2)

5\lambda

This is incorrect option

Option 3)

\frac{\text{5}} {2}\lambda

This is incorrect option

Option 4)

3\lambda

This is correct option

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