Energy of an electron in an excited hydrogen | vedclass

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(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} 	ext{ eV}$.Given $E_n = -3.4 	ext{ eV}$,

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$2.11 	imes 10^{-34} 	ext{ J s}$

Vedclass · Answer

(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} 	ext{ eV}$.Given $E_n = -3.4 	ext{ eV}$,we have $-\frac{13.6}{n^2} = -3.4$,which implies $n^2 = \frac{13.6}{3.4} = 4$,so $n = 2$.According to Bohr's postulate,the angular momentum $L$ is given by $L = \frac{nh}{2\pi}$.Substituting $n = 2$ and $h = 6.626 	imes 10^{-34} 	ext{ J s}$,we get $L = \frac{2h}{2\pi} = \frac{h}{\pi}$.$L = \frac{6.626 	imes 10^{-34}}{3.1416} \approx 2.11 	imes 10^{-34} 	ext{ J s}$.

Energy of an electron in an excited hydrogen atom is $-3.4 \text{ eV}$. Its angular momentum will be: $(h = 6.626 \times 10^{-34} \text{ J s})$

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