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(C) The energy of an incident photon is given by $E = \frac{12400}{\lambda (	ext{in } \mathring A)} \ eV$. For $\lambda_1 = 2500 \ \mathring A$,$E_1

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$6$

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(C) The energy of an incident photon is given by $E = \frac{12400}{\lambda (	ext{in } \mathring A)} \ eV$. For $\lambda_1 = 2500 \ \mathring A$,$E_1 = \frac{12400}{2500} = 4.96 \ eV$. The maximum kinetic energy is $K_1 = E_1 - \Phi = 4.96 - 2 = 2.96 \ eV$. Thus,the stopping potential $V_{s1} = 2.96 \ V$. For $\lambda_2 = 5000 \ \mathring A$,$E_2 = \frac{12400}{5000} = 2.48 \ eV$. The maximum kinetic energy is $K_2 = E_2 - \Phi = 2.48 - 2 = 0.48 \ eV$. Thus,the stopping potential $V_{s2} = 0.48 \ V$. The ratio of stopping potentials is $\frac{V_{s1}}{V_{s2}} = \frac{2.96}{0.48} \approx 6.16$. Rounding to the nearest integer option,the ratio is $6:1$.

If the wavelengths of incident radiation are $2500 \ \mathring A$ and $5000 \ \mathring A$ respectively,and the work function of the metal surface is $2 \ eV$,find the approximate ratio of the stopping potentials for the emitted photoelectrons. (in $:1$)

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