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

Hint: 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{align}
 &\Rightarrow e{{V}_{0}}=hv-W \\
 &\Rightarrow hv=e{{V}_{0}}+W \\
\end{align} $
Substituting the values we get;
 $ \begin{align}
 &\Rightarrow hv=10eV+2.75eV \\
 &\Rightarrow hv=12.75eV\text{ }......................\text{ (1)} \\
\end{align} $
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{align}
 &\Rightarrow hv={{E}_{n}}-{{E}_{1}} \\
 &\Rightarrow hv=\dfrac{-13.6}{{{n}^{2}}}-\left( \dfrac{-13.6}{{{1}^{2}}} \right) \\
\end{align} $
(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{align}
 &\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{align} $
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.