Question

Calculate the energy gap between the first energy state and the third excited state of a hydrogen like atom, given that the energy of the first excited state is $-3.4\text{eV}$.

• A. 3.4 eV
• B. 12 eV
• C. 12.8 eV
• D. 1.51 eV

Answer 1

The correct option is C 12.8 eV

We know that the energy of $n^{th}$ level in a Hydrogen like atom is given by $=-13.6\left(\frac{Z^2}{n^2}\right)$
According to the question, energy of the first excited state i.e. $n=2$ is $-3.4\text{eV}$.
So, $-13.6\frac{Z^2}{n^2}=-3.4$ (where, $n=2$)
$\Rightarrow -13.6\times \frac{Z^2}{4}=-3.4$
$\Rightarrow Z=1$

Energy of the first state,
$=\left(E_1\right)=-13.6\frac{(1{)}^2}{(1{)}^2}=-13.6\text{eV}$

Energy of the third excited state i.e., $n=4\left(E_4\right)$
$=-13.6\frac{(1{)}^2}{(4{)}^2}=-0.85\text{eV}$

Hence, the energy gap between the third excited state and the first energy state $=E_4-E_1=(-0.85)-(-13.6)=12.75\text{eV}\approx 12.8\text{eV}$