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Electrons are emitted from the cathode of a photocell of negligible work function when photons of wavelength \lambda  are incident on it. Derive the expression for the de Broglie wavelength of the electrons emitted in terms of the wavelength of the incident light.

 

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The maximum Kinetic energy of the emitted electron
\frac{1}{2}mv^{2}= h\vartheta -\phi _{0}
given \phi _{0} is negligible.
\therefore \frac{1}{2}mv^{2}= \frac{hc}{\lambda }
v= \sqrt{\frac{2hc}{\lambda m}}
De Broglie wavelength

= \frac{h}{mv}
\lambda_{d} =\frac{h}{m}\sqrt{\frac{\lambda m}{hc}}
    = \sqrt{\frac{h\lambda }{mc}}
\lambda _{d} is the De Broglie wavelength
and \lambda is the wavelength of the photon.

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Safeer PP

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