JEE Main 2024 — 5 April, Shift 2 — Question 51

Shortest, $\frac{\mathrm{hc}}{\lambda}=-13.6\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$

$\lambda \downarrow \mathrm{E} \uparrow ; \frac{\mathrm{hc}}{\lambda_{0}}=-13.6(1)$

Balmer Series :

$\longrightarrow$ n=3 $n \mathrm{n}=2$ $\frac{\mathrm{hc}}{\lambda_{1}}=-13.6\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$

$\frac{\mathrm{hc}}{\lambda_{1}}=-13.6\left(\frac{1}{4}-\frac{1}{9}\right)$ $\frac{\mathrm{hc}}{\lambda_{1}}=-13.6 \times\left(\frac{5}{36}\right)$

$\Rightarrow \frac{-13.6 \lambda_{0}}{\lambda_{1}}=-13.6 \times \frac{5}{36}$ $\lambda_{1}=\frac{\lambda_{0} \times 36}{5}=\frac{915 \times 36}{5}=6588$