The total energy of an electron in the first excited state of hydrogen atom is about $-3.4 \mathrm{eV}$. Its…

The total energy of an electron in the first excited state of hydrogen atom is about $-3.4 \mathrm{eV}$. Its kinetic energy in this state is:
  1. $3.4 \mathrm{eV}$
  2. $6.8 \mathrm{eV}$
  3. $-3.4 \mathrm{eV}$
  4. $-6.8 \mathrm{eV}$

Solution

$\mathrm{K} . \mathrm{E} .=\left|\frac{1}{2} P . E\right|$ But P.E is negative $\begin{aligned} \therefore \text { Total energy } & =\left|\frac{1}{2} P \cdot E\right|-P \cdot E \\ & =-\frac{P \cdot E}{2}=-3.4 \mathrm{eV} \\ \mathrm{K} . \mathrm{E} & =+3.4 \mathrm{eV} \end{aligned}$

Asked in: NEET 2005

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