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When photons of energy h\nu are incident on the surface of photosensitive material of work function h\nu _{0} , then -

Option: 1

the kinetic energy of all emitted electrons is h\nu_{0}


Option: 2

the kinetic energy of all emitted electrons is h\left ( \nu-\nu_{0} \right )


Option: 3

the kinetic energy of all fastest electrons is h\left ( \nu-\nu_{0} \right )


Option: 4

the kinetic energy of all emitted electrons is h\nu


Answers (1)

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\begin{aligned} & \frac{1}{2} \mathrm{mv}_{\max }^2=\mathrm{hv}-\mathrm{h} \mathrm{v}_0 \\ & =\mathrm{h}\left(\mathrm{v}-\mathrm{v}_0\right) \end{aligned}
This is Einstein's equation of the photoelectric effect.
 

Posted by

Ritika Harsh

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