1. Energy levels for hydrogen-like species

For a hydrogen-like ion, $E_n = -\frac{Z^2 R_H}{n^2}$ where:

• $Z$ = atomic number
• $R_H = 2.18 \times 10^{-18}\,\text{J}$

For $\mathrm{Li}^{2+}$, we have $Z=3$.

So, $E_n = -\frac{9R_H}{n^2}$

2. Energy required for excitation from $n$ to $(n+1)$

The absorbed energy is $\Delta E = E_{n+1} - E_n$

Using the formula, $\Delta E = -\frac{9R_H}{(n+1)^2} - \left(-\frac{9R_H}{n^2}\right)$ $\Delta E = 9R_H\left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right)$

Given: $\Delta E = 1.47 \times 10^{-17}\,\text{J}$

Substitute $R_H = 2.18 \times 10^{-18}\,\text{J}$: $1.47 \times 10^{-17} = 9(2.18 \times 10^{-18})\left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right)$

3. Simplify

First, $9 \times 2.18 \times 10^{-18} = 19.62 \times 10^{-18} = 1.962 \times 10^{-17}$

So, $1.47 \times 10^{-17} = 1.962 \times 10^{-17}\left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right)$

Divide both sides by $1.962 \times 10^{-17}$: $\frac{1.47}{1.962} = \frac{1}{n^2} - \frac{1}{(n+1)^2}$

$0.749 \approx \frac{1}{n^2} - \frac{1}{(n+1)^2}$

4. Test integer values of $n$

For $n=1$: $\frac{1}{1^2} - \frac{1}{2^2} = 1 - \frac14 = \frac34 = 0.75$

This matches the required value.

Hence, $n=1$

5. Verification by direct energy calculation

For excitation from $1$ to $2$ in $\mathrm{Li}^{2+}$: $\Delta E = 9R_H\left(1-\frac14\right)=9R_H\cdot\frac34$ $\Delta E = \frac{27}{4}R_H = 6.75(2.18\times10^{-18})$ $\Delta E = 1.4715\times10^{-17}\,\text{J}$

This agrees with the given value $1.47\times10^{-17}\,\text{J}$.

Therefore, the required integer is: $\boxed{1}$