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Q.

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 ……

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answer is 4.

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Detailed Solution

KEmax=10eV

φ=2.75eV

Total incident energy 

E=φ+KEmax=12.75eV

∴ Energy is released when electron jumps from the excited state n to the ground state. 

∵E4−E1={−0.85−(−13.6)} eV=12.75 eV

∴ value of n=4

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