Angular momentum of an electron in a hydrogen | vedclass

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(C) According to Bohr's quantization condition, the angular momentum $L$ of an electron in an orbit is given by $L = \frac{nh}{2\pi}$.Given that $L =

Vedclass · Accepted Answer

-$0.38$

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(C) According to Bohr's quantization condition, the angular momentum $L$ of an electron in an orbit is given by $L = \frac{nh}{2\pi}$.Given that $L = \frac{3h}{\pi}$, we equate the two expressions: $\frac{nh}{2\pi} = \frac{3h}{\pi}$.Solving for $n$, we get $n = 6$.The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula $E_n = -\frac{13.6}{n^2} 	ext{ eV}$.Substituting $n = 6$ into the formula: $E_6 = -\frac{13.6}{6^2} = -\frac{13.6}{36}$.Calculating the value: $E_6 \approx -0.377 	ext{ eV}$, which rounds to $-0.38 	ext{ eV}$.

Angular momentum of an electron in a hydrogen atom is $\frac{3h}{\pi}$, then the energy of the electron is . . . . . . eV.

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