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In the hydrogen spectrum, $\mathrm{ \lambda}$ be the wavelength of first transition line of Lyman series. The wavelength difference will be $\mathrm{ "a\lambda^{\text {" }}}$ between the wavelength of $\mathrm{ 3^{\text {rd }}}$ transition line of Paschen series and that of $\mathrm{2^{\text {nd }}}$ transition line of Balmer series where $\mathrm{a=}$ _____________.
Option: 1
5

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

First transition line of lyman series

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

$\mathrm{\lambda =\frac{4 }{3 R} \rightarrow \text { (1) }}$
Third transition line of paschen

$\mathrm{\frac{1}{\lambda_1} =R\left(\frac{1}{3^2}-\frac{1}{6^{2}}\right) }$

$\mathrm{\frac{1}{\lambda_1} =\frac{3 R}{36}}$

$\mathrm{\lambda _{1}=\frac{12}{R}}\rightarrow (2)$

2nd transition line of balmer series,

$\mathrm{\frac{1}{\lambda_2} =R\left(\frac{1}{2^2}-\frac{1}{4^2}\right) }$

$\mathrm{\lambda_2 =\frac{16}{3 R} \rightarrow(3) }$

$\mathrm{\lambda_1-\lambda_2 =\frac{12}{R}-\frac{16}{3 R} }$

$\mathrm{a \lambda =\frac{20}{3 R} \text { (Given) } }$

from eq (1)

$\mathrm{\text { a }\left(\frac{4}{3 R}\right) =\frac{20}{3 R} }$

$\mathrm{a=5}$

The value of $\mathrm{a=5}$

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