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A photoelectric surface is illuminated successively by monochromatic light of wavelengths \lambda and \frac{\lambda }{2} . If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, Among the following the work function of the surface is :

Option: 1

\frac{h c}{3 \lambda}


Option: 2

\frac{h c}{2 \lambda}


Option: 3

\frac{h c}{\lambda}


Option: 4

\frac{3 h c}{\lambda}


Answers (1)

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The correct option is 2)  \frac{h c}{2 \lambda}


In the first case, \quad \mathrm{E}_0=\frac{\mathrm{hc}}{\lambda}-\phi\, \, \, \, \, .....(1)


And in 2 \mathrm{nd}$ case, $3 \mathrm{E}_0=\frac{\mathrm{hc}}{\lambda / 2}-\phi\, \, \, \, \, \, ......(2)


Dividing 2 by 1


\begin{aligned} & (2) /(1) \Rightarrow \frac{3}{1}=\frac{\frac{2 h c}{\lambda}-\phi}{\frac{\mathrm{hc}}{\lambda}-\phi} \\ & \text { or } \frac{3 h c}{\lambda}-3 \phi=\frac{2 h c}{\lambda}-\phi \\ & \text { or } \phi=\frac{\text { hc }}{2 \lambda} \end{aligned}

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manish painkra

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