Question

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

• A. $13.6\text{eV}$
• B. $27.2\text{eV}$
• C. $40.8\text{eV}$
• D. $54.4\text{eV}$

Answer 1

The correct option is D $54.4\text{eV}$

Separation energy is the energy required to remove an electron from the given state to infinity. It is the energy difference between the given energy state and energy state at infinity.
For $B{e}^{3+}$, the separation energy is equal to $E_{\infty }-E_2$
$=0-(-13.6\frac{Z^2}{n^2})=+13.6\frac{Z^2}{n^2}$
$=13.6\times \frac{4^2}{2^2}=54.4\text{eV}$