Q98 / 200 · NEET 2024 · Physics

Match List I with List II. \begin{array}{|l|l|l|l|}\hline & {|c|}{\begin{array}{c} List I \\(Spectral Lines of \\Hydrogen for \\transitions from)\end{array}} & & {|c|}{\begin{array}{l} List II \\(Wavelengths (nm))\end{array}} \\\hline A. & n_2=3 to m_1=2 & I. & 410.2 \\\hline B. & n_2=4 to m_1=2 & II. & 434.1 \\\hline C. & n_2=5 to n_1=2 & III. & 656.3 \\\hline D. & n_2=6 to n_1=2 & IV. & 486.1 \\\hline\end{array} Choose the correct answer from the options given below:

  • Correct answer
    A
    A-III, B-IV, C-III, D-I
  • B
    A-IV, B-III, C-I, D-II
  • C
    A-I, B-II, C-III, D-IV
  • D
    A-II, B-I, C-IV, D-III

Solution

Energy difference $\Delta E=\frac{h c}{\lambda}$ $\begin{aligned} & \therefore \lambda \propto \frac{1}{\Delta E} \\ & (\Delta E)_{6-2}\gt(\Delta E)_{5-2}\gt(\Delta E)_{4-2}\gt(\Delta E)_{3-2} \\ & \lambda_{6-2} \lt \lambda_{5-2} \lt \lambda_{4-2} \lt \lambda_{3-2} \end{aligned}$ A-III, B-IV, C-II, D-I

Hence, the correct answer is option: 'A'.

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