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When a metal surface is illuminated by light of wavelengths 400nm and 250nm, the maximum velocities of the photoelectrons ejected are v and 2v respectively. The work function of the metal is -
(h = Planck's constant, c = velocity of light in air)

Option: 1

2hc\times10^{6}J


Option: 2

1.5hc\times10^{6}J


Option: 3

hc\times10^{6}J


Option: 4

0.5hc\times10^{6}J


Answers (1)

best_answer

\begin{aligned} & \frac{1}{2} \mathrm{mv}_{\max }^2=\mathrm{K} \cdot \mathrm{E}_{\cdot \max }=\frac{\mathrm{hc}}{\lambda}-\mathrm{W} \\ & \frac{1}{2} \mathrm{mv}_1^2=\frac{\mathrm{hc}}{\lambda_1}-\mathrm{W} \end{aligned}               ...................\left ( 1 \right )
\frac{1}{2} \mathrm{mv}_2^2=\frac{\mathrm{hc}}{\lambda_2}-\mathrm{W}                                              ...................\left ( 2 \right )
\begin{aligned} & \frac{\mathrm{eq}^{\mathrm{n}}(1)}{\mathrm{eq}^{\mathrm{n}}(2)} \Rightarrow\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}}=\left(\frac{\mathrm{v}}{2 \mathrm{v}}\right)^2 \\ & \frac{\mathrm{hc}}{\lambda_2}-\mathrm{W}=4\left(\frac{\mathrm{hc}}{\lambda_1}-\mathrm{W}\right) \\ & 3 \mathrm{~W}=\frac{4 \mathrm{hc}}{\lambda_1}-\frac{\mathrm{hc}}{\lambda_2} \end{aligned}
\\ \mathrm{W}=\frac{\mathrm{hc}}{3}\left[\frac{4}{\lambda_1}-\frac{1}{\lambda_2}\right]=\frac{\mathrm{hc}}{3}\left[\frac{4}{400 \times 10^{-9}}-\frac{1}{250 \times 10^{-9}}\right] \\ \mathrm{W}=\frac{\mathrm{hc}}{3} \times 10^9 \times \frac{150}{100 \times 250}=0.5 \mathrm{hc} \times 10^6 \mathrm{~J}

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Nehul

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