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2.9 A photon of wavelength 4 × 10 m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate (iii) the velocity of the photoelectron (1\ \textup{eV} = 1.6020 \times 10^{-19} \textup{J}).

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The photon is having a wavelength of 4\times 10^{-7}m strikes on a metal surface, where the work function of the metal being is 2.13\ eV.

From the previous part, we have the Kinetic Energy (K.E.):

K.E. = \frac{1}{2}mv^2 = 0.97\ eV = 0.97\times1.602\times10^{-19}J

\Rightarrow \frac{1}{2}\left ( 9.11\times10^{-31}kg \right )\times v^2 = 0.97\times1.602\times10^{-19}J

\because \left ( mass\ of\ an\ electron = 9.11\times10^{-31}kg \right )

\Rightarrow v^2 = 0.341\times10^{12} = 34.1\times10^{10}m/s

\Rightarrow v = 5.84\times10^5 m/s

Posted by

Divya Prakash Singh

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