NEETPhysicsAtoms and NucleiMCQ+4 / −1
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, then the value of n is
- A2
- B3
- C4
- D5
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Correct answer: C
KEmax = 10eV
= 2.75 eV
Total incident energy
E = + KEmax = 12.75 eV
Energy is released when electron jumps from
the excited state n to the ground state.
E4 – E1 = {– 0.85 – (–13.6) ev}
= 12.75eV
value of n = 4
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