Question

Calculate the ratio of energies of $H{e}^{+}$ for $1^{st}$ & $2^{nd}$ excited state.

• A. $9:4$
• B. $4:9$
• C. $1:4$
• D. $4:1$

Answer 1

The correct option is A $9:4$

The value n for $1^{st}$ and $2^{nd}$ excited state is 2 and 3 respectively
From Bohr's theory, if an electron of an hydrogen like species, with atomic number ${}^{\prime }{Z}^{\prime }$, orbits in orbit ${}^{\prime }{n}^{\prime }$ with energy ${}^{\prime }{E}_n^{\prime }$ then, $E_n\propto \frac{Z^2}{n^2}$
So,
Ratio of energies for $1^{st}$excited state$(n=2)$ and $2^{nd}$ excited state $(n=3)$ for $H{e}^{+}$ is
$\frac{E_2}{E_3}=\frac{4\times 9}{4\times 4}=\frac{9}{4}$