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  • The electric field at a point associated with a light wave is given by  <img alt="\mathrm{E=200\left[\sin \left(6 \times 10^{15}\right) t+\sin \left(9 \times 10^{15}\right) t\right]

The electric field at a point associated with a light wave is given by 

\mathrm{E=200\left[\sin \left(6 \times 10^{15}\right) t+\sin \left(9 \times 10^{15}\right) t\right] \mathrm{Vm}^{-1}}

\mathrm{\text { Given : } h=4.14 \times 10^{-15} \mathrm{eVs}}

If this light falls on a metal surface having a work function of \mathrm{2.50 \mathrm{eV}}, the maximum kinetic energy of the photoelectrons will be 

Option: 1

\mathrm{1.90 \; \mathrm{eV}}


Option: 2

\mathrm{3.27 \, \, \mathrm{eV}}


Option: 3

\mathrm{3.60\, \, \mathrm{eV}}


Option: 4

\mathrm{3.42\, \, \mathrm{eV}}


Answers (1)

best_answer

\mathrm{E=200\left[\sin \left(6 \times 10^{15} t\right)+\sin \left(9 \times 10^{15} t\right)\right]}

\mathrm{Freqneny\; of\; EM \;waves\; are}

\mathrm{=\frac{6}{2 \pi} \times 10^{15} \text { and } \frac{9}{2 \pi} \times 10^{15}}

\text{Energy of one photon of these waves}

\mathrm{E_{1}=h f_{1} =4.14 \times 10^{-15} \times \frac{6}{2 \pi} \times 10^{15}} \\

\mathrm{=3.95 \mathrm{eV}} \\

\mathrm{E_{2}=h f_{2} =4.14 \times 10^{-15} \times \frac{9}{2\pi} \times 10^{15}} \\

\mathrm{=5.93 \mathrm{eV}}

\text{Energy of maximum kinetic}

\mathrm{E =W_{0}+K E_{\text {max }}} \\

\mathrm{K E_{m ax} =5.93-2.50} \\

\mathrm{=3.43 \mathrm{eV}}

Hence the correct option is 4

Posted by

Devendra Khairwa

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