MCQ
If energy of the electron in hydrogen atom in some excited state is $-3.4\,eV,$ then what will be its angular momentum
  • A
    $1.8 \times 10^{-30} \, kg\,m^2\,s^{-1}$
  • $2.1  \times 10^{-34} \, kg\,m^2\,s^{-1}$
  • C
    $9.2  \times 10^{-37} \, kg\,m^2\,s^{-1}$
  • D
    $1.2  \times 10^{-32} \, kg\,m^2\,s^{-1}$

Answer

Correct option: B.
$2.1  \times 10^{-34} \, kg\,m^2\,s^{-1}$
b
Given, $E=-3.4 \mathrm{eV}$

Energy, $E=\frac{-13.6 \mathrm{eV}}{n^{2}}$

$n^{2}=\frac{-13.6 \mathrm{eV}}{-3.4 \mathrm{eV}}$

$n^{2}=4$

$n=2$

From Bohr's theory, the angular momentum id given by $L=\frac{n h}{2 \pi}$

$L=\frac{(2) h}{2 \pi}$

$L=\frac{h}{\pi}$

The correct option is $B$.

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