The energy of an electron in an orbit of the Bohr's atom is $- 0.04E_{0}eV$ where $E_{0}$ is the gro

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The energy of an electron in an orbit of the Bohr's atom is $- 0.04E_{0}eV$ where $E_{0}$ is the ground state energy. If L is the angular momentum of the electron in this orbit and h is the Planck's constant, then $\frac{2\pi	ext{ }L}{	ext{ }h}$ is $\_\_\_\_$ :

Accepted Answer

Angular momentum $L = \frac{nh}{2\pi}$$$n = \frac{2\pi	ext{ }L}{	ext{ }h} $$Energy $E = - \frac{13.6}{n^{2}}.z^{2}$$$\begin{matrix} & E \Rightarrow - \frac{E_{0}}{n^{2}} = - 0.04E_{0} \ & n^{2} = 25,n = 5 \end{matrix}$$

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