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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = \frac{hc}{\lambda} - \phi$,where $\phi$

Vedclass · Accepted Answer

$4/3 \, eV$

Vedclass · Answer

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.For $\lambda_1 = 3100 \mathring A$,$K_1 = \frac{1}{2} m v_1^2 = \frac{12400}{3100} - \phi = 4 - \phi$.For $\lambda_2 = 6200 \mathring A$,$K_2 = \frac{1}{2} m v_2^2 = \frac{12400}{6200} - \phi = 2 - \phi$.Given the ratio of speeds $v_1 : v_2 = 2 : 1$,the ratio of kinetic energies is $K_1 : K_2 = v_1^2 : v_2^2 = 4 : 1$.Therefore,$\frac{4 - \phi}{2 - \phi} = \frac{4}{1}$.$4 - \phi = 4(2 - \phi) = 8 - 4\phi$.$3\phi = 4 \implies \phi = 4/3 \, eV$.

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