Question

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. What is the value of n?
(A) 3
(B) 4
(C) 5
(D) 2

Answer 1

When an electron absorbs energy, it jumps to a higher orbital. This is called an excited state. An electron in an excited state can release energy and fall to a lower state. When it does, the electron releases a photon of electromagnetic energy. To find out the value of n, we will use the Einstein’s photoelectric equation i.e.
$\Rightarrow e{{V}_{0}}=hv-W$

Complete step by step answer:
Here, Stopping potential, ${{V}_{0}}=10V$
Work function, W = 2.75 eV
According to Einstein’s photoelectric equation
$\begin{aligned} &\Rightarrow e{{V}_{0}}=hv-W \\\ &\Rightarrow hv=e{{V}_{0}}+W \\\ \end{aligned}$
Substituting the values we get;
$\begin{aligned} &\Rightarrow hv=10eV+2.75eV \\\ &\Rightarrow hv=12.75eV\text{ }......................\text{ (1)} \\\ \end{aligned}$
Now,
When an electron in the hydrogen atom makes a transition from excited state n to the ground state (n=1), then the frequency (v) of the emitted photon is given by:
$\begin{aligned} &\Rightarrow hv={{E}_{n}}-{{E}_{1}} \\\ &\Rightarrow hv=\dfrac{-13.6}{{{n}^{2}}}-\left( \dfrac{-13.6}{{{1}^{2}}} \right) \\\ \end{aligned}$
(Because for hydrogen atom, Bohr’s atomic energy in nth orbit is given by $~\text{ }{{E}_{n}}=\dfrac{-13.6}{{{n}^{2}}}eV$ ).
According to the given problem,
$\Rightarrow hv=\dfrac{-13.6}{{{n}^{2}}}+13.6$
Using equation (1),
$\begin{aligned} &\Rightarrow \dfrac{-13.6}{{{n}^{2}}}+13.6=12.75 \\\ &\Rightarrow \dfrac{-13.6}{{{n}^{2}}}=-0.85 \\\ &\Rightarrow {{n}^{2}}=\dfrac{13.6}{0.85} \\\ &\Rightarrow {{n}^{2}}=16 \\\ &\Rightarrow n=4 \\\ \end{aligned}$
Therefore option (B) is the correct answer.

Note:
We can also calculate the de-Broglie wavelength from this by using Rydberg formula i.e.
$\Rightarrow \dfrac{1}{\lambda }=\overline{v}=R\left[ \dfrac{1}{{{n}_{f}}^{2}}-\dfrac{1}{{{n}_{i}}^{2}} \right]$
Where $\overline{v}$ is the wave number,$R$ is the Rydberg constant, ${{n}_{f}}$ and ${{n}_{i}}$ are the final and initial number of orbits from which emission of electrons takes place.