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Two identical metal plates show photoelectric effect. Light of wavelength \lambda _{A} falls on plate A and \lambda _{B} falls on plate B. \lambda _{A}= 2\lambda _{B}. The maximum K.E. of the photoelectrons are K_{A} and K_{B} respectively.
Which one of the following is true ?

Option: 1

2K_{A}= K_{B}


Option: 2

K_{A}= 2K_{B}


Option: 3

K_{A}< K_{B}/2


Option: 4

K_{A}> 2K_{B}


Answers (1)

\mathrm{K}_{\mathrm{A}}=\frac{\mathrm{hc}}{\lambda_{\mathrm{A}}}-\phi=\frac{\mathrm{hc}}{2 \lambda_{\mathrm{B}}}-\phi                     .................\left ( i \right )
\mathrm{K}_{\mathrm{B}}=\frac{\mathrm{hc}}{\lambda_{\mathrm{B}}}-\phi \Rightarrow \frac{\mathrm{hc}}{\lambda_{\mathrm{B}}} \quad=\mathrm{K}_{\mathrm{B}}+\phi         ................\left ( ii \right )
From eq^{n}\left ( i \right ) & \left ( ii \right )
\begin{aligned} & \mathrm{K}_{\mathrm{A}}=\frac{1}{2}\left(\mathrm{~K}_{\mathrm{B}}+\phi\right)=\frac{1}{2} \mathrm{~K}_{\mathrm{B}}-\frac{\phi}{2} \\ & \mathrm{~K}_{\mathrm{A}}<\frac{1}{2} \mathrm{~K}_{\mathrm{B}} \end{aligned}

Posted by

Sumit Saini

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