The stopping potential for photoelectrons fro | vedclass

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Question

(D) According to Einstein's photoelectric equation,the maximum kinetic energy of photoelectrons is given by $K_{max} = h v - \phi_{0}$,where $\phi_{0}

Vedclass · Accepted Answer

$v_{1}+\frac{e}{h}(V_{2}-V_{1})$

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(D) According to Einstein's photoelectric equation,the maximum kinetic energy of photoelectrons is given by $K_{max} = h v - \phi_{0}$,where $\phi_{0}$ is the work function of the metal.Since $K_{max} = e V_{s}$,where $V_{s}$ is the stopping potential,we have $e V_{s} = h v - \phi_{0}$.For the first case: $e V_{1} = h v_{1} - \phi_{0}$  ---$(i)$For the second case: $e V_{2} = h v_{2} - \phi_{0}$  ---(ii)Subtracting equation $(i)$ from equation (ii):$e V_{2} - e V_{1} = (h v_{2} - \phi_{0}) - (h v_{1} - \phi_{0})$$e(V_{2} - V_{1}) = h(v_{2} - v_{1})$$v_{2} - v_{1} = \frac{e}{h}(V_{2} - V_{1})$$v_{2} = v_{1} + \frac{e}{h}(V_{2} - V_{1})$

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