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Solve this problem - Atomic Structure - JEE Main

   If \lambda _{0} and \lambda be the threshold wavelength and wavelength of incident light, the velocity of photoelectron ejected from the metal surface is :

  • Option 1)

    \sqrt{\frac{2h}{m}\left ( \lambda _{0}-\lambda \right )}

  • Option 2)

    \sqrt{\frac{2hc}{m}\left ( \lambda _{0}-\lambda \right )}

  • Option 3)

    \sqrt{\frac{2hc}{m}\left ( \frac{\lambda _{0}-\lambda }{\lambda \lambda _{0}} \right )}

  • Option 4)

    \sqrt{\frac{2h}{m}\left ( \frac{1}{\lambda _{0}}-\frac{1}{\lambda } \right )}

 
Answers (2)
77 Views
N neha

As discussed in the concept

Photoelectric Effect -

\frac{1}{2}mu^{2}= hv-hv_{0}

- wherein

where

m is the mass of the electron

u is the velocity associated with the ejected electron.

h is plank’s constant.

v is frequency of photon,

v0 is threshold frequency of metal.

 

 v^{2} = \frac{2hc}{m}\left ( \frac{1}{\lambda } - \frac{1}{\lambda _{0}} \right )

v =\sqrt{ \frac{2hc}{m}\left ( \frac{1}{\lambda } - \frac{1}{\lambda _{0}} \right )}

 

Hence the correct option is 3

 


Option 1)

\sqrt{\frac{2h}{m}\left ( \lambda _{0}-\lambda \right )}

Option 2)

\sqrt{\frac{2hc}{m}\left ( \lambda _{0}-\lambda \right )}

Option 3)

\sqrt{\frac{2hc}{m}\left ( \frac{\lambda _{0}-\lambda }{\lambda \lambda _{0}} \right )}

Option 4)

\sqrt{\frac{2h}{m}\left ( \frac{1}{\lambda _{0}}-\frac{1}{\lambda } \right )}

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