JEE Mains · Physics · STD 12 - Atoms
- JEE Mains 2020
- Hard
- Numerical
In the line spectra of hydrogen atom, difference between the largest and the shortest wavelengths of the Lyman series is \(304\,\mathring {A}.\) The corresponding difference for the Paschan series in \(\mathring {A}\) is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(10553\)
Step-by-step Solution
Detailed explanation
\(\lambda=\frac{ c }{\left(\frac{1}{ n _{1}^{2}}-\frac{1}{ n _{2}^{2}}\right)}\)
for lyman series
\(\lambda_{1}=\frac{ c }{\frac{1}{1^{2}}-\frac{1}{\infty^{2}}}= c ( n =\infty\) to \(n =1)\)
\(\lambda_{2}=\frac{c}{\frac{1}{1^{2}}-\frac{1}{2^{2}}}=\frac{4 c}{3}(n=2\) to \(n=1)\)
\(\Delta \lambda=\lambda_{2}-\lambda_{1}=\frac{ c }{3}=304 A \Rightarrow c =912 A\)
for paschen series
\(\lambda_{1}=\frac{ c }{\frac{1}{3^{2}}-\frac{1}{\infty^{2}}}=9 c ( n =\infty\) to \(n =3)\)
\(\lambda_{2}=\frac{ c }{\frac{1}{3^{2}}-\frac{1}{4^{2}}}=\frac{144 c }{7}( n =4\) to \(n =3)\)
\(\Delta \lambda=\lambda_{2}-\lambda_{1}=\frac{144 c }{7}-9 c =\frac{81 c }{7}=\frac{81 \times 912}{7}\)
\(=10553.14 \mathring {A}\)
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