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  4. Angular momentum of an electron in hydrogen atom is (3h/2π) (h is the Planck’s constant). The K.E. of the electron is

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Q. Angular momentum of an electron in hydrogen atom is $\frac{3h}{2\pi}$ (h is the Planck’s constant). The K.E. of the electron is

KCETKCET 2020

Solution:

Angular momentum,
$L =\frac{ nh }{2 \pi}=\frac{3 h }{2 \pi} $
$ given$
also $ n =3$
$\therefore E =\frac{-1.36}{ n ^{2}} eV$ $E =\frac{-13.6}{(3)^{2}} eV$
$E =\frac{-13.6}{9} eV$
$E = -1.57 \,eV$