A photon emitted during the de-excitation of electron from a state n to the first excited state in a hydrogen atom, irradiates a metallic cathode of work function 2\; eV, in a photo cell, with a stopping potential of 0.55\; V. Obtain the value of the quantum number of the state n.

 

 

Answers (1)

From einstein photoelectric equation, We know that

        hv=\phi _{o}+eV_{s}

Where, \phi is the work function

            eV_{s} is stopping potential

So,       h\nu=2+0.55

            h\nu=0.55\; eV

We have, E_{n}=\frac{-13.6}{n^{2}}

Then, the energy difference, \Delta E=-3.4-(-2.55)eV=-0.85\; eV

Therefore,         \frac{-13.6}{n^{2}}=0.85

                \therefore \; \; n=4

The value of the quantum number of the state n is 4.

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