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  • Find the maximum  of photo electrons emitted from the surface of lithium <img alt="\left ( \phi =2.39eV \right )" src="

Find the maximum KE of photo electrons emitted from the surface of lithium \left ( \phi =2.39eV \right ) when exposed with E=E_{0}\left ( 1+cos6\times 10^{14} t \right ) cos 3.6 \times 10^{15}t

Option: 1

0.37 eV


Option: 2

0.1 eV


Option: 3

0.02 eV


Option: 4

0.06 eV


Answers (1)

best_answer

For maximum KE , frequency should be maximum  \omega = 4.2 \times 10^{5} s^{-1}
\left ( KE \right )_{max} = \frac{h\omega }{2\pi }-\phi=0.37 eV

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Gaurav

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