Question

The electron in the hydrogen atom jumps 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 $5.1 eV$, the stopping potential is estimated to be (the energy of the electron in $n^{\text {th }}$ state $\left.E_{n}=\frac{-13.6}{n^{2}} e V\right)$

• A. 5.1 V
• B. 12.1 V
• C. 17.2 V
• D. 7 V

Answer 1

Answer: (D)

Energy released when electron in the atom jumps from excited state $(n=3)$ to ground state $(n=1)$ is
$E=h v=E_{3}-E_{1}=\frac{-13.6}{3^{2}}-\left(\frac{-13.6}{1^{2}}\right) $
$=\frac{-13.6}{9}+13.6=12.1\, eV$
Therefore, stopping potential
$e V_{0}=h v-\phi_{0} $
$=12.1-5.1 \left[\because \text { work function } \phi_{0}=5.1\right] $
$ V_{0}=7 \,V$