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(A) According to Einstein's photoelectric equation: $K_{max} = \frac{hc}{\lambda} - \phi = eV,$ where $\phi = \frac{hc}{\lambda_0}$ is the work functi

Vedclass · Accepted Answer

$9$

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(A) According to Einstein's photoelectric equation: $K_{max} = \frac{hc}{\lambda} - \phi = eV,$ where $\phi = \frac{hc}{\lambda_0}$ is the work function.For wavelength $\lambda$: $\frac{hc}{\lambda} = \frac{hc}{\lambda_0} + eV$ ... $(i)$For wavelength $3\lambda$: $\frac{hc}{3\lambda} = \frac{hc}{\lambda_0} + \frac{eV}{4}$ ... (ii)Multiply equation (ii) by $4$: $\frac{4hc}{3\lambda} = \frac{4hc}{\lambda_0} + eV$ ... (iii)Equating the expressions for $eV$ from $(i)$ and (iii):$\frac{hc}{\lambda} - \frac{hc}{\lambda_0} = \frac{4hc}{3\lambda} - \frac{4hc}{\lambda_0}$Rearranging terms: $\frac{4hc}{\lambda_0} - \frac{hc}{\lambda_0} = \frac{4hc}{3\lambda} - \frac{hc}{\lambda}$$\frac{3hc}{\lambda_0} = \frac{hc}{3\lambda}$$\frac{3}{\lambda_0} = \frac{1}{3\lambda} \implies \lambda_0 = 9\lambda$Given $\lambda_0 = n\lambda,$ therefore $n = 9.$

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