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JEE Mains · Physics · STD 12 - Atoms
  • JEE Mains 2019
  • Hard
  • MCQ

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....\(nm\)

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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 \((\mathrm{2})\)

\(\frac{\lambda}{660}=\frac{5 \times 16}{36 \times 3}\)

\(\lambda=\frac{4400}{9}=488.88=488.9\, \mathrm{nm}\)

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