# Solve this following

Question:

In the line spectra of hydrogen atom, difference between the largest and the shortest wavelengths of the Lyman series is $304 \AA$. The corresponding difference for the Paschan series in $A$ is :

Solution:

$\lambda=\frac{\mathrm{c}}{\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)}$

for lyman series

$\lambda_{1}=\frac{\mathrm{c}}{\frac{1}{1^{2}}-\frac{1}{\infty^{2}}}=\mathrm{c}(\mathrm{n}=\infty$ to $\mathrm{n}=1)$

$\lambda_{2}=\frac{\mathrm{c}}{\frac{1}{1^{2}}-\frac{1}{2^{2}}}=\frac{4 \mathrm{c}}{3}(\mathrm{n}=2$ to $\mathrm{n}=1)$

$\Delta \lambda=\lambda_{2}-\lambda_{1}=\frac{c}{3}=304 \AA \Rightarrow c=912 A$

for paschen series

$\lambda_{1}=\frac{\mathrm{c}}{\frac{1}{3^{2}}-\frac{1}{\infty^{2}}}=9 \mathrm{c} \quad(\mathrm{n}=\infty$ to $\mathrm{n}=3)$

$\lambda_{2}=\frac{\mathrm{c}}{\frac{1}{3^{2}}-\frac{1}{4^{2}}}=\frac{144 \mathrm{c}}{7}(\mathrm{n}=4$ to $\mathrm{n}=3)$

$\Delta \lambda=\lambda_{2}-\lambda_{1}=\frac{144 \mathrm{c}}{7}-9 \mathrm{c}=\frac{81 \mathrm{c}}{7}=\frac{81 \times 912}{7}$

$=10553.14 A$