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 \(Wave

Question

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:NEET 2024

Options:

Answer:

A

Solution:

Energy difference $Δ E = \frac{h c}{λ}$ $∴ λ \propto \frac{1}{Δ E} (Δ E)_{6 - 2} > (Δ E)_{5 - 2} > (Δ E)_{4 - 2} > (Δ E)_{3 - 2} λ_{6 - 2} < λ_{5 - 2} < λ_{4 - 2} < λ_{3 - 2}$ A-III, B-IV, C-II, D-I

TestHub · Accepted Answer

Energy difference Δ E = λ h c ∴ λ ∝ Δ E 1 ( Δ E ) 6 − 2 > ( Δ E ) 5 − 2 > ( Δ E ) 4 − 2 > ( Δ E ) 3 − 2 λ 6 − 2 < λ 5 − 2 < λ 4 − 2 < λ 3 − 2 A-III, B-IV, C-II, D-I

Physics - Modern Physics Question with Solution | TestHub

Options:

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