Question

Ratio of frequency of revolution of electron in second excited state of $H{e}^{+}$ and first excited state of $H$ is $\frac{2^x}{3^y}$. Then the value of $x+y$ is:

Answer 1

$\nu \propto \frac{Z^2}{n^3}$

For $H{e}^{+},Z=2,n=3$

For $H^{+},Z=1,n=2$

Ratio of the frequency of revolution of the electron in second excited state of $H{e}^{+}$ and first excited state of $H$ is

$\frac{{\nu }_{H{e}^{+}}}{{\nu }_H}=\frac{(\frac{Z^2}{n^3}{)}_{H{e}^{+}}}{(\frac{Z^2}{n^3}{)}_H}$

$\frac{{\nu }_{H{e}^{+}}}{{\nu }_H}=\frac{\frac{2^2}{3^3}}{\frac{1^2}{2^3}}=\frac{2^5}{3^3}=\frac{2^x}{2^y}$

Thus, $x=5$ and $y=3$

$x+y=5+3=8$