Question

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

Answer 1

$\lambda=\frac{c}{\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)}$
for lyman series
$\lambda_{1}=\frac{c}{\frac{1}{1^{2}}-\frac{1}{\infty^{2}}}=c(n=\infty$ to $n=1$
$\lambda_{2}=\frac{c}{\frac{1}{1^{2}}-\frac{1}{2^{2}}}=\frac{4 c}{3}(n=2$ to $n=1)$
$\Delta \lambda=\lambda_{2}-\lambda_{1}=\frac{c}{3}=304 \mathring{A}$
$ \Rightarrow c=912 \mathring{A}$
for paschen series
$\lambda_{1}=\frac{c}{\frac{1}{3^{2}}-\frac{1}{\infty^{2}}}=9 c(n=\infty$ to $n=3)$
$\lambda_{2}=\frac{c}{\frac{1}{3^{2}}-\frac{1}{4^{2}}}=\frac{144 c}{7}(n=4$ to $n=3)$
$\Delta \lambda=\lambda_{2}-\lambda_{1}=\frac{144 c}{7}-9 c$
$=\frac{81 c}{7}=\frac{81 \times 912}{7}$
$=10553.14 \mathring{A}$