The angular momentum of an electron in a stat | vedclass

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(B) The angular momentum is given by $mvr = \frac{nh}{2\pi}$.Given angular momentum $= \frac{3h}{\pi} = \frac{6h}{2\pi}$,so $n = 6$.For the radius of

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$6.348 \ \mathring{A}, -5.45 	imes 10^{-19} \ J$

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(B) The angular momentum is given by $mvr = \frac{nh}{2\pi}$.Given angular momentum $= \frac{3h}{\pi} = \frac{6h}{2\pi}$,so $n = 6$.For the radius of the orbit,$r_n = a_0 	imes \frac{n^2}{Z}$,where $a_0 = 0.529 \ \mathring{A}$.$r_6 = 0.529 \ \mathring{A} 	imes \frac{6^2}{3} = 0.529 	imes 12 = 6.348 \ \mathring{A}$.For the energy of the stationary state,$E_n = -2.18 	imes 10^{-18} \ J 	imes \frac{Z^2}{n^2}$.$E_6 = -2.18 	imes 10^{-18} \ J 	imes \frac{3^2}{6^2} = -2.18 	imes 10^{-18} 	imes \frac{9}{36} = -2.18 	imes 10^{-18} 	imes \frac{1}{4} = -5.45 	imes 10^{-19} \ J$.

$\lambda \ (nm)$	$500, 450, 400$
$v \times 10^{-5} \ (cm \ s^{-1})$	$2.55, 4.35, 5.35$

The angular momentum of an electron in a stationary state of $Li^{2+} (Z=3)$ is $\frac{3h}{\pi}$. The radius and energy of that stationary state are respectively:

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