Question

Electrons revolving around the nucleus in $n_1^{th}$ orbit of an atom have atomic number $Z_1$ and in the $n_2^{th}$ orbit of other atom have atomic number $Z_2$, then:
[where P = linear momentum, L = angular momentum, f = frequency of revolution and K.E. = kinetic energy]

• A. $\frac{L_1}{L_2}=\frac{n_1}{n_2}$
• B. $\frac{P_1}{P_2}=\frac{Z_1}{Z_2}\times \frac{n_2}{n_1}$
• C. $\frac{f_1}{f_2}=\left(\frac{Z_2}{Z_1}{)}^2\right(\frac{n_1}{n_2}{)}^3$
• D. $\frac{(K.E{)}_1}{(K.E.{)}_2}=(\frac{Z_1}{Z_2}.\frac{n_2}{n_1}{)}^2$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $\frac{L_1}{L_2}=\frac{n_1}{n_2}$
B $\frac{P_1}{P_2}=\frac{Z_1}{Z_2}\times \frac{n_2}{n_1}$
D $\frac{(K.E{)}_1}{(K.E.{)}_2}=(\frac{Z_1}{Z_2}.\frac{n_2}{n_1}{)}^2$

According to the Bohr model, angular momentum, $L=mvr=nh/2\pi$.

So, $\frac{L_1}{L_2}=\frac{n_1}{n_2}$ and $K.E=-\frac{13.6Z^2}{n^2}$

$⟹\frac{(K.E{)}_1}{(K.E.{)}_2}=(\frac{Z_1}{Z_2}.\frac{n_2}{n_1}{)}^2$

Also, as $P=\sqrt{2mK.E}$

$⟹\frac{P_1}{P_2}=\frac{Z_1}{Z_2}\times \frac{n_2}{n_1}$