Solution:
Formula: \(eV_0 = E - \Phi_0\), where \(E = \frac{hc}{\lambda} = \frac{12400}{2000} = 6.2\text{ eV}\). Thus, the stopping potential \(V_0 = 6.2 - 4.5 = 1.7\text{ V}\), requiring an applied potential of \(-1.7\text{ V}\).
The potential difference that must be applied to stop the fastest moving photoelectrons emitted by a metal surface, having work function \(4.5\text{ eV}\), when ultraviolet light of \(2000A^\circ\) falls on it, will be
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