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2.12    Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (ν0) and work function (W0) of the metal.

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Given the wavelength of radiation is 6800\ \AA.

Energy\ given = Work\ function + Kinetic\ energy.

But the electrons are emitted with zero velocity from a metal surface when it is exposed to radiation. That means the kinetic energy will be zero.

So, the Threshold frequency \nu_{o} will be:

\nu_{o} = \frac{c}{\lambda_{o}} = \frac{3.0\times10^8 m/s}{6800\times10^{-10}m} = 4.14\times10^{14}s^{-1}

and the Work function will be:

W_{o} = h\nu_{o} = 6.626\times10^{-34}Js \times 4.14\times10^{14}s^{-1} = 2.92\times10^{-19}J

Posted by

Divya Prakash Singh

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