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The surface of a certain metal was illuminated alternately with light of wavelength \lambda_1 and \lambda_2\left(\lambda_1<\lambda_2\right) and it was found that the corresponding maximum velocities of the photoelectrons differ by a factor of \eta. Then the work function of the metal is given by:

Option: 1

hc \, \frac{\eta^2\left[1-\frac{\lambda_2}{\lambda_1}\right]}{\lambda_2\left(\eta^2-1\right)}


Option: 2

hc \, \frac{\eta^2\left[1-\frac{2 \lambda_2}{\lambda_1}\right]}{\lambda_2\left(\eta^2-1\right)}


Option: 3

hc \, \frac{\eta^2\left[1-\frac{\lambda_2}{2 \lambda_1}\right]}{\lambda_2\left(\eta^2-1\right)}


Option: 4

None of the Above


Answers (1)

best_answer

If \mathrm{v_1} and \mathrm{v_2} be the maximum velocities of photoelectrons corresponding to wavelengths \lambda_1 and \lambda_2

\text { then, } \quad \frac{1}{2} \mathrm{mv}_1^2=\frac{\mathrm{hc}}{\lambda_1}-\mathrm{W}

\frac{1}{2} \mathrm{mv}_2^2=\frac{\mathrm{hc}}{\lambda_2}-\mathrm{W} \text {; with } \mathrm{v}_1=\eta \mathrm{v}_2

\text { Thus, }\left(\frac{\mathrm{v}_1}{\mathrm{v}_2}\right)^2=\frac{\frac{\mathrm{hc}}{\lambda_1}-\mathrm{W}}{\frac{\mathrm{hc}}{\lambda_2}-\mathrm{W}}=\eta^2

\text { or, } \quad \eta^2\left(\frac{\mathrm{hc}}{\lambda_2}-\mathrm{W}\right)=\left(\frac{\mathrm{hc}}{\lambda_1}-\mathrm{W}\right)

Simplifying this we get

W=\frac{h c \eta^2\left(1-\frac{\lambda_2}{\lambda_1}\right)}{\lambda_2\left(\eta^2-1\right)}

Posted by

Pankaj Sanodiya

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