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

Answers (1)

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

Satyajeet Kumar

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