Question

In hydrogen and hydrogen-like atoms, the ratio of $E_{4n}-E_{2n}$ and $E_{2n}-E_n$ varies with atomic number $z$ and principal quantum number $n$ as

• A. $\frac{z^2}{n^2}$
• B. $\frac{z^4}{n^4}$
• C. $\frac{z}{n}$
• D. None of these

Answer 1

Answer: (D)

None of these

$E\left(n\right)=13.6Z^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
$\frac{E\left(4n\right)-E\left(2n\right)}{E\left(2n\right)-E\left(n\right)}=\frac{\frac{1}{(4n{)}^2}-\frac{1}{(2n{)}^2}}{\frac{1}{(2n{)}^2}-\frac{1}{n^2}}$
$\frac{E\left(4n\right)-E\left(2n\right)}{E\left(2n\right)-E\left(n\right)}=\frac{\frac{1}{16}-\frac{1}{4}}{\frac{1}{4}-1}=\frac{1}{4}$