The angular momentum of an electron in the hy | vedclass

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Question

(B) According to Bohr's postulate,the angular momentum $L$ of an electron in the $n^{th}$ orbit is given by $L = \frac{nh}{2\pi}$.Given $L = \frac{3h}

Vedclass · Accepted Answer

$1.51$

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(B) According to Bohr's postulate,the angular momentum $L$ of an electron in the $n^{th}$ orbit is given by $L = \frac{nh}{2\pi}$.Given $L = \frac{3h}{2\pi}$,comparing the two expressions,we get $n = 3$.The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} 	ext{ eV}$.For $n = 3$,$E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.51 	ext{ eV}$.The kinetic energy $K$ of an electron in an orbit is equal to the magnitude of its total energy,so $K = |E_n| = 1.51 	ext{ eV}$.

The angular momentum of an electron in the hydrogen atom is $\frac{3h}{2\pi}$. Here $h$ is Planck's constant. The kinetic energy of this electron is.....$eV$

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