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Emission and Absorption Spectra

The electrons...

Question

The electrons in the hydrogen atom lumps from excited state $(n = 3)$ to its ground state $(n = 1)$ and the photons thus emitted irradiate a photosensitive material. If the work function of the material is $5.1 e V$ . Calculate the stopping potential (the energy of the electron in $n^{t h}$ state $E_{n} = - e V)$ .

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Solution

To find the radiating energy
$E = 13.6 (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
$= 13.6 (\frac{1}{1} - \frac{1}{9})$
$= 13.6 (\frac{8}{9}) e V$
$= 12.08$ $e^{-} V$
radiated energy = work function + $K E$
$= 12.08$ $e^{-} V$ - $5.1$ $e^{-} V$ = $K E$
$K E = 6.98 e^{-} V$
Stopping portental $= \frac{K E}{e} =$ = $= \frac{6.98 e V}{e} = 6.98 V$

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