Atoms and Nuclei Part 2

Since, we know that, $\;\frac{1}{\lambda} = R z^{2} \left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)$......(i) where, $\lambda =$ wavelength of light emitted. $R =$ Rydberg's constant, $Z =$ atomic number, $n_{1} =$ principal quantum number of lower energy level and $\;\; n_{2} =$ principal quantum number of higher energy level. Therefore, for $1 \text{st}$ spectral line of Balmer series, $n_{1} = 2$ and $n_{2} = 3$ $\;\frac{1}{\lambda_{1}} = R z^{2} \left( \frac{1}{2^{2}} - \frac{1}{3^{2}} \right)$ $\Rightarrow \;\; \frac{1}{\lambda_{1}} = R z^{2} \left( \;\frac{5}{36} \right)$...(ii) Similarly, for 3rd spectral line, $n_{1} = 2$ and $n_{2} = 5$ $\;\frac{1}{\lambda_{3}} = R z^{2} \left( \frac{1}{2^{2}} - \frac{1}{5^{2}} \right)$ $\Rightarrow \;\; \frac{1}{\lambda_{3}} = R z^{2} \left( \;\frac{21}{100} \right)$......(iii) Now, dividing Eq. (iii) by Eq. (ii), we get $\; \frac{ \;\frac{1}{\lambda_{3}} }{ \frac{1}{\lambda_{1}} } = \frac{ R z^{2} \left( \frac{21}{100} \right) }{ R z^{2} \left( \;\frac{5}{36} \right) } \Rightarrow \frac{\lambda_{1}}{\lambda_{3}} = \frac{21}{100} \times \frac{36}{5}$ $\Rightarrow \;\; \frac{\lambda_{1}}{\lambda_{3}} = 1.512 = 15.12 \times 10^{-1}$ Comparing with the given value in the question i.e., $x \times 10^{-1}$, the value of $x = 15$.