The energy of an electron in the 3^{rd} orbit | vedclass

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Question

(D) According to Bohr's theory,the energy of an electron in the $n^{th}$ orbit is given by $E_n = -\frac{Z^2 R_H}{n^2}$,which implies $E_n \propto -\f

Vedclass · Accepted Answer

$-9E$

Vedclass · Answer

(D) According to Bohr's theory,the energy of an electron in the $n^{th}$ orbit is given by $E_n = -\frac{Z^2 R_H}{n^2}$,which implies $E_n \propto -\frac{1}{n^2}$.Given that the energy in the $3^{rd}$ orbit is $E_3 = -E$.We know $E_n = \frac{k}{n^2}$ where $k$ is a constant.For $n=3$,$E_3 = \frac{k}{3^2} = \frac{k}{9} = -E$,so $k = -9E$.For the $1^{st}$ orbit $(n=1)$,$E_1 = \frac{k}{1^2} = k$.Substituting the value of $k$,we get $E_1 = -9E$.Therefore,the correct option is $D$.

The energy of an electron in the $3^{rd}$ orbit of a hydrogenic atom is $-E$. The energy of an electron in the first orbit will be

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