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(B) Given,$E = -3.4 \ eV$. The energy of an electron in the $n^{th}$ orbit is given by $E = -\frac{13.6 \ eV}{n^2}$. Substituting the value: $-3.4 = -

Vedclass · Accepted Answer

$2.1 	imes 10^{-34} \ kg \ m^2 \ s^{-1}$

Vedclass · Answer

(B) Given,$E = -3.4 \ eV$. The energy of an electron in the $n^{th}$ orbit is given by $E = -\frac{13.6 \ eV}{n^2}$. Substituting the value: $-3.4 = -\frac{13.6}{n^2} \implies n^2 = \frac{13.6}{3.4} = 4 \implies n = 2$. According to Bohr's postulate,the angular momentum $L$ is given by $L = \frac{nh}{2\pi}$. For $n = 2$,$L = \frac{2h}{2\pi} = \frac{h}{\pi}$. Using $h = 6.626 	imes 10^{-34} \ J \ s$ and $\pi \approx 3.14$: $L = \frac{6.626 	imes 10^{-34}}{3.14} \approx 2.11 	imes 10^{-34} \ kg \ m^2 \ s^{-1}$. Thus,the correct option is $B$.

If the energy of the electron in a hydrogen atom in some excited state is $-3.4 \ eV,$ then what will be its angular momentum?

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