Question

In a hydrogen like ion, the energy difference between the 2\textsuperscript{nd} excitation energy state and ground is 108.8 eV. The atomic number of the ion is:

Answer 1

3

Answer 2

For a hydrogen-like ion, the energy of the nth level is

$$E_n = -\dfrac{13.6\,Z^2}{n^2}\; \text{eV}.$$

Here, the "2nd excitation energy state" refers to the second excited state, which is $n = 3$. The energy difference between the ground state ($n = 1$) and $n = 3$ is

$$\Delta E = E_1 - E_3 = \left(-\dfrac{13.6\,Z^2}{1^2}\right) - \left(-\dfrac{13.6\,Z^2}{3^2}\right)= 13.6\,Z^2\left(1 - \dfrac{1}{9}\right) = 13.6\,Z^2\left(\dfrac{8}{9}\right).$$

Setting $\Delta E = 108.8$ eV:

$$13.6\,Z^2\left(\dfrac{8}{9}\right) = 108.8.$$

Solving for $Z^2$:

$$Z^2 = \frac{108.8 \times 9}{13.6 \times 8} = \frac{979.2}{108.8} = 9,$$

so,

$$Z = 3.$$