If $\lambda$1 and $\lambda$2 are the wavelengths of the third member of Lyman and first member of the Paschen series respectively, then the value of $\lambda$1 : $\lambda$2 is :
Solution
For Lyman series
<br><br>n<sub>1</sub> = 1, n<sub>2</sub> = 4
<br><br>$${1 \over {{\lambda _1}}} = R\left[ {{1 \over {{1^2}}} - {1 \over {{4^2}}}} \right]$$
<br><br>For paschen series
<br><br>n<sub>1</sub> = 3, n<sub>2</sub> = 4
<br><br>$${1 \over {{\lambda _2}}} = R\left[ {{1 \over {{3^2}}} - {1 \over {{4^2}}}} \right]$$<br><br>$\therefore$ $${{{\lambda _1}} \over {{\lambda _2}}} = {{\left[ {{1 \over 9} - {1 \over {16}}} \right]} \over {\left[ {1 - {1 \over {16}}} \right]}} = {7 \over {9 \times 15}}$$<br><br>$\Rightarrow$ ${{{\lambda _1}} \over {{\lambda _2}}} = {7 \over {135}}$
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Bohr's Model of Hydrogen Atom
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