Solution:
Formula: \(E = K_{max} + \Phi_0 = 8 + 4.09 = 12.09\text{ eV}\). For hydrogen atom transition, \(13.6 \left(1 - \frac{1}{n^2}\right) = 12.09\). Solving this gives \(n = 3\).
An electron in the hydrogen atom jumps from state \(n\) to the ground state. The wavelength so emitted illuminates a photo-sensitive material having work function \(4.09\text{ eV}\). If the stopping potential of the photoelectrons is \(8\text{ V}\), the value of \(n\) is
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