Atomic Structure - Result Question 101
####77. Consider the hydrogen atom to be proton embedded in a cavity of radius $a _0$ (Bohr’s radius) whose charge is neutralised by the addition of an electron to the cavity in vacuum, infinitely slowly. Estimate the average total energy of an electron in its ground state in a hydrogen atom as the work done in the above neutralisation process. Also, if the magnitude of the average kinetic energy is half the magnitude of the average potential energy, find the average potential energy.
(1996, 2M)
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Answer:
Correct Answer: 77. (10)
Solution:
- The work done in the given neutralisation process is
$$ \begin{gathered} \quad W=-\int _{a _0}^{\infty} F d r \text { and } F=\frac{e^{2}}{4 \pi \varepsilon _0 r^{2}} \ \Rightarrow \quad W=\frac{e^{2}}{4 \pi \varepsilon _0}\left[\frac{1}{r}\right] _{a _0}^{\infty}=-\frac{e^{2}}{4 \pi \varepsilon _0 r}=\text { Total energy }(E) \end{gathered} $$
Now, if ’ $V$ ’ is magnitude of potential energy, then according to given information, kinetic energy $\left(E _k\right)$ is $V / 2$. Therefore,
$$ \begin{aligned} E & =-V+\frac{V}{2} \quad(PE \text { is always negative }) \ & =-\frac{V}{2} \ \Rightarrow \quad V & =-2 E=\frac{-e^{2}}{2 \pi \varepsilon _0 r} \end{aligned} $$
