In the hydrogen spectrum, $\lambda$ be the wavelength of first transition line of Lyman series. The wavelength difference will be "a$\lambda$'' between the wavelength of $3^{\text {rd }}$ transition line of Paschen series and that of $2^{\text {nd }}$ transition line of Balmer series where $\mathrm{a}=$ ___________.
Answer (integer)
5
Solution
<p>${1 \over \lambda } = {R_H}\left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right)$</p>
<p>$${1 \over {{\lambda _3}}} = {R_H}\left( {{1 \over {{3^2}}} - {1 \over {{6^2}}}} \right)$$</p>
<p>$${1 \over {{\lambda _2}}} = {R_H}\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right)$$</p>
<p>$\therefore$ ${\lambda _3} - {\lambda _2} = a\lambda$</p>
<p>$a = 5$</p>
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Bohr's Model of Hydrogen Atom
This question is part of PrepWiser's free JEE Main question bank. 184 more solved questions on Atoms and Nuclei are available — start with the harder ones if your accuracy is >70%.