Question

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

• A. $Z^2/n^2$
• B. $Z^4/n^4$
• C. $Z/n$
• D. $Z^0/n^0$

Answer 1

The correct option is D $Z^0/n^0$

$E_n=\frac{Z^2{E}_1}{n^2}$ .....................(1)

$E_1$ is ground state energy of hydrogen atom.

So for getting $E_{4n}$ just replace $n$ by $4n$ in the eqaution-1

Thus we get $E_{4n}=E_1/16n^2$

Similarly $E_{2n}=E_1/4n^2$

So the ratio will be $\frac{E_{4n}-E_{2n}}{E_{2n}-E_n}=\frac{4n^2(1-4)}{16n^2(1-4)}=\frac{1}{4}\propto Z^0/n^0$

Option D is correct.