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

Answer: 10553

$\lambda =\frac{c}{\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)}$
for lyman series
${\lambda }_1=\frac{c}{\frac{1}{1^2}-\frac{1}{{\infty }^2}}=c(n=\infty$ to $n=1$
${\lambda }_2=\frac{c}{\frac{1}{1^2}-\frac{1}{2^2}}=\frac{4c}{3}(n=2$ to $n=1)$
$\Delta \lambda ={\lambda }_2-{\lambda }_1=\frac{c}{3}=304\mathring{A}$
$\Rightarrow c=912\mathring{A}$
for paschen series
${\lambda }_1=\frac{c}{\frac{1}{3^2}-\frac{1}{{\infty }^2}}=9c(n=\infty$ to $n=3)$
${\lambda }_2=\frac{c}{\frac{1}{3^2}-\frac{1}{4^2}}=\frac{144c}{7}(n=4$ to $n=3)$
$\Delta \lambda ={\lambda }_2-{\lambda }_1=\frac{144c}{7}-9c$
$=\frac{81c}{7}=\frac{81\times 912}{7}$
$=10553.14\mathring{A}$