In the line spectra of hydrogen atom,

Question:

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

Solution:

(10553.14)

From Bohr's formula for hydrogen atom,

$\frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$

$R=1.097 \times 10^{7} \mathrm{~m}^{-1}$

For Lyman series :

$\frac{1}{\lambda_{\min .}}=R(1)=R \quad \because n_{2}=\infty$ and $n_{1}=1$

$\frac{1}{\lambda_{\max }}=R\left\{1-\frac{1}{4}\right\}=\frac{3 R}{4} \quad \because n_{1}=2, n_{1}=1$

$\therefore \lambda_{\max }-\lambda_{\min .}=\frac{4}{3 R}-\frac{1}{R}=\frac{1}{3 R}=304$ (Given)

For Paschen series :

$\lambda_{\min .}^{\prime}=R\left(\frac{1}{9}\right)$ and $\lambda_{\max .}^{\prime}=R\left(\frac{1}{9}-\frac{1}{16}\right)=\frac{7 R}{16 \times 9}$

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

or, $\lambda_{\max .}^{\prime}-\lambda_{\min .}^{\prime}=\frac{81}{7 R}=\frac{81 \times 3}{7 \times 3 R}=\frac{81 \times 3}{7} \times 304$

$\left(\because \frac{1}{3 R}=304 Aright)$

$\therefore$ For Pachen series, $\lambda_{\max .}^{\prime}-\lambda_{\min .}^{\prime}=10553.14$

 

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