Question

What is the seperation energy (in eV) for $B{e}^{3+}$ in the first excited state in eV?

• A. $13.6eV$
• B. $27.2eV$
• C. $40.8eV$
• D. $54.4eV$

Answer 1

The correct option is D $54.4eV$

Separation Energy= $13.6\times \frac{Z^2}{n^2}$

For $Be$, $Z=4$

For $1^{st}$ excited state $n=2$

$\therefore 13.6\times \frac{4^2}{2^2}=54.4eV$