The maximum KE of the photoelectron is found to be 6.63\times10^{-19}J, when the metal is irradiated with a radiation of frequency 2\times10^{15} H_{2} . The threshold frequency of the metal is about

  • Option 1)

    1\times10^{15}s^{-1}

  • Option 2)

    2\times10^{15}s^{-1}

  • Option 3)

    3\times10^{15}s^{-1}

  • Option 4)

    1.5\times10^{15}s^{-1}

 

Answers (1)
P Plabita

As learnt in

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.

 

 \frac{1}{2}mu^{2} = maximum KE

V_{0} = threshold frequency

\therefore  we have

6.63\times 10^{-19} = 6.63\times 10^{-34} (2\times 10^{15}hz-v_{0})

V_{0} = 1\times 10^{15} s^{-1}

 


Option 1)

1\times10^{15}s^{-1}

This solution is correct.

Option 2)

2\times10^{15}s^{-1}

This solution is incorrect.

Option 3)

3\times10^{15}s^{-1}

This solution is incorrect.

Option 4)

1.5\times10^{15}s^{-1}

This solution is incorrect.

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