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

Vedclass · Accepted Answer

$4$

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(A) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $K_{max} = eV_s = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function and $\lambda_t$ is the threshold wavelength such that $\phi = \frac{hc}{\lambda_t}$.For the first case: $e(3V_0) = \frac{hc}{\lambda} - \phi$  --- $(i)$For the second case: $eV_0 = \frac{hc}{2\lambda} - \phi$  --- (ii)Multiply equation (ii) by $3$: $3eV_0 = \frac{3hc}{2\lambda} - 3\phi$  --- (iii)Equating $(i)$ and (iii): $\frac{hc}{\lambda} - \phi = \frac{3hc}{2\lambda} - 3\phi$$2\phi = \frac{3hc}{2\lambda} - \frac{hc}{\lambda} = \frac{hc}{2\lambda}$$\phi = \frac{hc}{4\lambda}$Since $\phi = \frac{hc}{\lambda_t}$,we have $\frac{hc}{\lambda_t} = \frac{hc}{4\lambda}$,which implies $\lambda_t = 4\lambda$.Thus,the value of $n$ is $4$.

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