JEE Mains · Physics · STD 12 - 12. atoms
Taking the wavelength of first Balmer line in hydrogen spectrum (\( n = 3\) to \(n = 2\) ) as \(660\,nm ,\) the wavelength of the \(2^{nd}\) Balmer line ( \(n =4\) to \(n = 2\) ) will be....\(nm\)
- A \(889.2\)
- B \(642.7\)
- C \(448.9\)
- D \(388.9\)
Answer & Solution
Correct Answer
(C) \(448.9\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{660}=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\frac{5 R}{36}\) .... \((1)\) \(\frac{1}{\lambda}=R\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)=\frac{3 \mathrm{R}}{16}\) .... \((2)\) Divide equation \((1)\) with \((\mathrm{2})\)…
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