Question

Ratio of radii of orbits in two different orbits of $H-$ atom is $4:9$, then possible ratio of energy for these two orbits is:

• A. $2:3$
• B. $3:2$
• C. $2:9$
• D. $9:4$

Answer 1

The correct option is D $9:4$

$r_n=\text{radius of}{n}^{th}\text{orbit}$
$E_n=\text{Energy of}{n}^{th}\text{orbit}$
We know,
$\frac{r_{n_1}}{r_{n_2}}=\frac{0.529\left(\frac{n_1^2}{1}\right)\stackrel{o}{A}}{0.529\times \left(\frac{n_2^2}{1}\right)\stackrel{o}{A}}\Rightarrow \frac{4}{9}=\frac{n_1^2}{n_2^2}\cdots \left(1\right)$
Now,
$\frac{E_{n_1}}{E_{n_2}}=\frac{-13.6\times \left(\frac{1}{n_1^2}\right)\text{eV}}{-13.6\times \left(\frac{1}{n_2^2}\right)\text{eV}}=\frac{n_2^2}{n_1^2}=\frac{9}{4}=9:4$