Question

In Hydrogen - like atoms, ratio of $E_{4n}-E_{2n}$ and $E_{2}n-E_n$ is proportional to

• A. $\frac{Z^2}{n^2}$
• B. $\frac{Z^4}{n^4}$
• C. $\frac{Z}{n}$
• D. $\frac{Z^0}{n^0}$

Answer 1

The correct option is D $\frac{Z^0}{n^0}$

For a H - like atom.
$E_n=\frac{Z^2}{n^2},E_1$ (where, $E_1=-13.6eV$)
$\Rightarrow E_{4n}-E_{2n}=\frac{Z^2{E}_1}{16n^2}-\frac{Z^2{E}_1}{4n^2}$
$=\frac{Z^2{E}_1}{n^2}(\frac{1}{16}-\frac{1}{4})$
$=\frac{Z^2{E}_1}{n^2}\left(\frac{-3}{16}\right)$
And $E_{2}n-E_n=\frac{Z^2{E}_1}{n_2}(\frac{1}{4}-\frac{1}{1})$
$=\frac{Z^2{E}_1}{n^2}\left(\frac{-3}{4}\right)$
Ratio $\frac{E_{4n}-E_{2n}}{E_{2n}-E_n}=\frac{4}{16}$
$=\frac{1}{4}$ = a constant.