Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

Match List-I with List-II.

List-I (Spectral Lines of Hydrogen for transitions from) A. n₂=3 to n₁=2 B. n₂=4 to n₁=2 C. n₂=5 to n₁=2 D. n₂=6 to n₁=2

List-II (Wavelength (nm)) I. 410.2 II. 434.1 III. 656.3 IV. 486.1

Choose the correct answer from the options given below:

A-III, B-IV, C-II, D-I

A-IV, B-III, C-I, D-II

A-I, B-II, C-III, D-IV

A-II, B-I, C-IV, D-III

Step-by-Step Solution

These transitions correspond to the Balmer series (transitions to $n_1=2$ ) in the hydrogen spectrum . The wavelength ( $\lambda$ ) is inversely proportional to the energy difference ( $\Delta E$ ) of the transition ( $\Delta E = hc/\lambda$ ).

Transition A ( $n_2=3 \to n_1=2$ ): This represents the smallest energy gap in the series, corresponding to the longest wavelength (Red line, $H_\alpha$ ). Matches III (656.3 nm).
Transition D ( $n_2=6 \to n_1=2$ ): This represents the largest energy gap among the given options, corresponding to the shortest wavelength (Violet line, $H_\delta$ ). Matches I (410.2 nm).
Transition B ( $n_2=4 \to n_1=2$ ): Intermediate energy ( $H_\beta$ ). Matches IV (486.1 nm).
Transition C ( $n_2=5 \to n_1=2$ ): Intermediate energy ( $H_\gamma$ ). Matches II (434.1 nm).

Therefore, the correct matching is A-III, B-IV, C-II, D-I.

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