Solution:
Solution
We know that, $$\Delta E=13.6 Z^2\left[\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right] eV$$
In first case, $$(10.2+17.0)=13.6 Z^2\left[\dfrac{1}{2^2}-\dfrac{1}{n^2}\right] . . .(1)$$
In second case, $$(4.25+5.95)=13.6 Z^2\left[\dfrac{1}{3^2}-\dfrac{1}{n^2}\right] . . . (2)$$
$$(1)/(2) \Rightarrow \dfrac{27.2}{10.2}=\dfrac{9(n^2-4)}{4(n^2-9)}$$
$$\Rightarrow 1.18=\dfrac{(n^2-4)}{(n^2-9)}$$
$$\Rightarrow n^2-4=1.18n^2-10.66$$
$$\Rightarrow 0.18n^2=6.66$$ or $$n^2 \sim 36$$
$$\Rightarrow n=6$$
Put the value of $$n$$ in (1), we get $$Z=3$$
