Question

The ratio of the energy of the electron in the ground state of hydrogen to the electron in first excited state of $B{e}^{3+}$ is

• A. 1 : 4
• B. 1 : 8
• C. 1 : 16
• D. 16 : 1

Answer 1

Answer: (A)

(1) $\to H(2)\to B{e}^{3+}$

$\frac{E_1}{E_2}=\frac{Z_1^2}{n_1^2}\times \frac{n_2^2}{Z_2^2}$

$\Rightarrow \frac{1}{1}\times \frac{4}{16}=\frac{1}{4}$