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An electron in the hydrogen atom jumps from excited state n to the ground state. The wavelength so emitted illuminates a photosensitive material having work function 2.75 eV. If the stopping potential of the photoelectron is 10 V, the value of n is

1. 3
2. 4
3. 5
4. 2

Correct answer: 5

Solution

The energy of the photon emitted during the transition is equal to the work function plus the kinetic energy of the ejected photoelectron. Using the photoelectric equation, we find the energy of the photon to be 2.75 eV + 10 eV = 12.75 eV. For the hydrogen atom, the energy difference between levels is given by E = 13.6(1 - 1/n²) eV. Solving for n, we find n = 5.

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