Question

In the line spectra of hydrogen atom, the difference between the largest and the shortest wavelength of the Lyman series is $304\mathring{A}$ The corresponding difference for the Paschen series (in $\mathring{A}$) is close to (upto two decimal places)

Answer 1

From, Bohr's atom model for hydrogen atom,

$\frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})[\because R=1.097\times {10}^7{\text{m}}^{-1}]$

For Lyman series:

$\frac{1}{{\lambda }_{min}}=R\left(1\right)=R[\because n_2=\infty \text{and}{n}_1=1]$
${\frac{1}{\lambda }}_{max}=R\{1-\frac{1}{4}\}=\frac{3R}{4}[\because n_2=2,n_1=1]$
$\therefore {\lambda }_{max}-{\lambda }_{min}=\frac{4}{3R}-\frac{1}{R}=\frac{1}{3R}=304\mathring{A}\text{( Given )}$

For Paschen series:

${\lambda }_{min}^{\prime }=R\left(\frac{1}{9}\right)[\because n_2=\infty \text{and}{n}_1=1]$

${\lambda }_{max}^{\prime }=R(\frac{1}{9}-\frac{1}{16})=\frac{7R}{16\times 9}[\because n_2=4,n_1=3]$

${\lambda }_{max}^{\prime }-{\lambda }_{min}^{\prime }=\frac{16\times 9}{7R}-\frac{9}{R}=\frac{81}{7R}$

$=\frac{81\times 3}{7\times 3R}=\frac{81\times 3}{7}\times 304=10553.14\mathring{A}[\because \frac{1}{3R}=304\mathring{A}]$