JEE Main 2026 — 22 January, Evening Shift — Question 32

Smallest wavelength of lyman $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{12}-\frac{1}{\infty^{2}}\right)$ $\mathrm{R}=\frac{1}{\lambda}=\frac{1}{91} \mathrm{~nm}^{-1}$ $\lambda_{\text {max }}$ for balmer series $\mathrm{n}_{1}=2 \rightarrow \mathrm{n}_{2}=3$ $\frac{1}{\lambda_{\mathrm{B}}}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{9}\right)$ $\frac{1}{\lambda_{\mathrm{B}}}=\frac{1}{91}\left(\frac{5}{36}\right)$ $\lambda_{\mathrm{B}}=\left(\frac{91 \times 36}{5}\right)=655.2 \mathrm{~nm}$ $\lambda_{\text {max }}$ paschen $\mathrm{n}_{1}=3 \rightarrow \mathrm{n}_{2}=4$ $\frac{1}{\lambda_{p}}=\frac{1}{91}\left(\frac{1}{3^{2}}-\frac{1}{4^{2}}\right)=\frac{1}{91} \times \frac{7}{144}$ $\lambda_{\mathrm{p}}=\left(\frac{91 \times 144}{7}\right)=1872 \mathrm{~nm}$ $\Delta \lambda=\lambda_{\mathrm{P}}-\lambda_{\mathrm{B}}=1872-655.2$ $\Delta \lambda=1216.8$ $\Delta \lambda \approx 1217$