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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 \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.

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

Similar Questions

Bird flu is caused by

Total internal reflection of light is possible when light enters from

If $x > 0, y > 0, z > 0, xy + yz + zx < 1$ and if $\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \pi$,then $x + y + z$ equals to

In the circuit shown in the figure,neglecting source resistance,the voltmeter and ammeter readings will respectively be:

Which of the following is the correct order of stability for alkali metal chlorides?

Vedclass Test Series

Exam Paper Generator

Online Exam Module