Angular momentum of an electron in a hydrogen | vedclass

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Question

(B) Given,angular momentum of electron in $H$-atom $= \frac{3h}{2\pi} \dots (i)$From Bohr's postulate,angular momentum $= \frac{nh}{2\pi} \dots (ii)$C

Vedclass · Accepted Answer

$1.51$

Vedclass · Answer

(B) Given,angular momentum of electron in $H$-atom $= \frac{3h}{2\pi} \dots (i)$From Bohr's postulate,angular momentum $= \frac{nh}{2\pi} \dots (ii)$Comparing equations $(i)$ and $(ii)$,we get $n = 3$.The kinetic energy $(KE)$ of an electron in a hydrogen atom is given by the formula:$KE = \frac{13.6 	imes Z^2}{n^2} 	ext{ eV}$For hydrogen atom,atomic number $Z = 1$. Substituting $Z = 1$ and $n = 3$:$KE = \frac{13.6 	imes 1^2}{3^2} 	ext{ eV}$$KE = \frac{13.6}{9} 	ext{ eV}$$KE = 1.51 	ext{ eV}$

Angular momentum of an electron in a hydrogen atom is $\frac{3h}{2\pi}$ ($h$ is the Planck's constant). The kinetic energy $(KE)$ of the electron is (in $\text{ eV}$)

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