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

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(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 \fr

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,the energy of the electron in the $3^{rd}$ orbit is $E_3 = -E$.Using the relation $E_n \propto \frac{1}{n^2}$,we have:$\frac{E_1}{E_3} = \frac{3^2}{1^2} = \frac{9}{1} = 9$.Therefore,$E_1 = 9 	imes E_3 = 9 	imes (-E) = -9E$.Thus,the energy in the first orbit is $-9E$,and option $(D)$ is correct.

List-$I$	List-$II$
$(A)$ Electron deficient hydride	$(I)$ $CH_4$
$(B)$ Electron precise hydride	$(II)$ $B_2H_6$
$(C)$ Electron rich hydride	$(III)$ $CaH_2$
$(D)$ Saline hydride	$(IV)$ $NiH_{0.6}$
	$(V)$ $PH_3$

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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Match the following:

List-$I$ List-$II$

$(A)$ Electron deficient hydride $(I)$ $CH_4$

$(B)$ Electron precise hydride $(II)$ $B_2H_6$

$(C)$ Electron rich hydride $(III)$ $CaH_2$

$(D)$ Saline hydride $(IV)$ $NiH_{0.6}$

$(V)$ $PH_3$

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