An electron in the hydrogen atom jumps from a higher energy state to the ground state. The emitted photon illuminates a photosensitive material having work function \( 2.75\ \mathrm{eV} \). If the stopping potential of the photoelectron is \( 10\ \mathrm{V} \), then the value of principal quantum number \( n \) characterizing the higher energy state is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 4 \)
The photon from the transition supplies the work function plus the maximum kinetic energy, which equals \( e \times \) stopping potential (Einstein’s photoelectric equation).
\( h\nu = \phi_{0} + eV_{0} = 2.75\ \mathrm{eV} + 10\ \mathrm{eV} = 12.75\ \mathrm{eV} \)
For a jump to the ground state of hydrogen:
\( \displaystyle 13.6\left(1 - \frac{1}{n^{2}}\right) = 12.75 \)
\( \displaystyle \frac{1}{n^{2}} = 1 - \frac{12.75}{13.6} = \frac{0.85}{13.6} = \frac{1}{16} \)
\( n = 4 \)