The energy of an electron in an excited hydro | vedclass

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Question

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = \frac{-13.6 \, eV}{n^2}$.Given $E_n = -3.4 \, eV$,we have $\

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$2.1 	imes 10^{-34}$

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(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = \frac{-13.6 \, eV}{n^2}$.Given $E_n = -3.4 \, eV$,we have $\frac{-13.6}{n^2} = -3.4$,which implies $n^2 = 4$,so $n = 2$.According to Bohr's postulate,the angular momentum $L$ is given by $L = \frac{nh}{2\pi}$.Substituting $n = 2$,we get $L = \frac{2h}{2\pi} = \frac{h}{\pi}$.Using $h = 6.626 	imes 10^{-34} \, J \cdot s$ and $\pi \approx 3.14$,we get $L = \frac{6.626 	imes 10^{-34}}{3.14} \approx 2.11 	imes 10^{-34} \, J \cdot s$.

The energy of an electron in an excited hydrogen atom is $-3.4 \, eV$. According to Bohr's theory,the angular momentum of this electron in $J \cdot s$ is:

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