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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

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Correct Answer

(A) \(10553\)

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\(\lambda=\frac{ c }{\left(\frac{1}{ n _{1}^{2}}-\frac{1}{ n _{2}^{2}}\right)}\)

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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