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Question

The total energy of an electron in the first excited state of the hydrogen atom is about $$-34eV$$.
(a) What is the kinetic energy of the electron in this state?
(b) What is the potential energy energy of the electron in this state?
(c) Which of the answers above would change if the choice of the zero of potential energy is changed?

Solution
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We know that K.E. and P.E of the electron in the $$n^{th}$$ energy state of hydrogen atom are respectively given by,
$$E_{k}=\dfrac{1}{4\pi \epsilon}\times\dfrac{e^{2}}{2r}$$
and $$E_{p}=-\dfrac{1}{4\pi \epsilon _{0}}\times \dfrac{e^{2}}{r_{n}}$$
Where, $$r_{n}=4\pi \epsilon _{0}\times\dfrac{n^{2}h^{2}}{4\pi ^{2}me^{2}}$$
The first excited state corresponds to $$n=2$$, level. Now, total energy of electron in $$n=2$$, state is given to be $$-3.4eV$$.
$$\therefore \dfrac{1}{4\pi \epsilon _{0}}\times\dfrac{e^{2}}{2r_{2}}+\left ( -\dfrac{1}{4\pi \epsilon _{0}}\times\dfrac{e^{2}}{r_{2}} \right )=-3.4eV$$
$$\Rightarrow \dfrac{1}{4\pi \epsilon _{0}}\times\dfrac{e^{2}}{2r_{2}}=3.4eV$$
(a) Therefore, K.E of the electron in the first excited state $$(n=2)$$ of hydrogen atom,
$$=\dfrac{1}{4\pi \epsilon _{0}}\times\dfrac{e^{2}}{2r_{2}}=3.4eV$$.
(b) $$E$$ of the electron in the first excited state $$(n=2)$$ of the hydrogen atom,
$$=\dfrac{1}{4\pi \epsilon _{0}}\times\dfrac{e^{2}}{2r_{2}}=-2\times\left ( \dfrac{1}{4\pi \epsilon _{0}}\times\dfrac{e^{2}}{2r_{2}} \right )$$
$$=-2\times3.4$$
$$=-6.8eV$$.
(c) If the zero of potential energy is chosen differently, K.E does not change. The P.E and total energy will change.

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