Q166 / 180 · KCET 2020 · Physics
Angular momentum of an electron in hydrogen atom is $\frac{3 h}{2 \pi}(h$ is the Planck's constant). The KE of the electron is
- A$4.35 \mathrm{eV}$
- Correct answerB$1.51 \mathrm{eV}$
- C$3.4 \mathrm{eV}$
- D$6.8 \mathrm{eV}$
Solution
Given, angular momentum of electron in
H-atom $=\frac{3 h}{2 \pi}...(i)$
From Bohr's postulate,
Angular momentum $=\frac{n h}{2 \pi}...(ii)$
Comparing Eqs. (i) and (ii), we get
$n=3$
The kinetic energy of an electron in hydrogen is given by
$\begin{aligned}
\mathrm{KE} &=\frac{13.6 \times Z^{2}}{n^{2}} \mathrm{eV} \\
&=\frac{13.6 \times 1}{3^{2}} \\
&=1.51 \mathrm{eV}
\end{aligned}$
$($ for $H, Z=1$ )
Hence, the correct answer is option: 'B'.