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

Vedclass · Accepted Answer

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

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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,$E_n = -2.18 	imes 10^{-18} \ J 	imes \frac{Z^2}{n^2}$.$E_6 = -2.18 	imes 10^{-18} \ J 	imes \frac{3^2}{6^2} = -2.18 	imes 10^{-18} 	imes \frac{9}{36} = -2.18 	imes 10^{-18} 	imes \frac{1}{4} = -5.45 	imes 10^{-19} \ J$.

$(P). \lambda_{\max }$ in Lyman series	$(i). \infty \rightarrow 1$
$(Q). v_{\min }$ in Balmer series	$(ii). 2 \rightarrow 1$
$(R). \lambda_{\min }$ in Lyman series	$(iii). 3 \rightarrow 2$
$(S). v_{\max }$ in Balmer series	$(iv). \infty \rightarrow 2$

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:

Similar Questions

Match the following for hydrogen atom $(Z=1)$ :-
$(P). \lambda_{\max }$ in Lyman series $(i). \infty \rightarrow 1$
$(Q). v_{\min }$ in Balmer series $(ii). 2 \rightarrow 1$
$(R). \lambda_{\min }$ in Lyman series $(iii). 3 \rightarrow 2$
$(S). v_{\max }$ in Balmer series $(iv). \infty \rightarrow 2$

What is the angular momentum of an electron when it revolves in orbits?

Light,a well-known form of energy,is treated as a form of matter,by saying that it consists of:

The energy of an electron in the first Bohr orbit of $H$ atom is $-13.6 \ eV$. The possible energy value$(s)$ of the excited state$(s)$ for electrons in Bohr orbits of the hydrogen atom is(are) $............ \ eV$.

The wave number of the first emission line in the Balmer series of $H$ atom spectrum is ($R =$ Rydberg constant).

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