JEE Mains · Physics · STD 12 - 12. atoms
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
- A \(10553\)
- B \(10560\)
- C \(10555\)
- D \(10165\)
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 \(n=1)\)…
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