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 (Wavelengths (nm)) I. 410.2 II. 434.1 III. 656.3 IV. 486.1 Choose the correct answer from the options given below :

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q2 – NEET-UG 2024

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Home Physics Structure of AtomQuestion #483

Match List I with List II.

\begin{array}{lcl} List I & List II \\ (Spectral Lines of Hydrogen for & (Wavelengths (nm)) \\ transitions from) \\ A . n_{2} = 3 to n_{1} = 2 & I. 410.2 \\ B. n_{2} = 4 to n_{1} = 2 & II. 434.1 \\ C. n_{2} = 5 to n_{1} = 2 & III. 656.3 \\ D. n_{2} = 6 to n_{1} = 2 & IV. 486.1 \end{array}

Choose the correct answer from the options given below :

Clue 1 of 3: Hydrogen Spectrum

The spectral lines of hydrogen are governed by transitions of electrons between different energy levels.

🧠 Full Solution Path

Step 1: Apply Rydberg Formula for Balmer Series✦ Active

The transitions are all to $n_{1} = 2$ , which corresponds to the Balmer series. The Rydberg formula for the wavelength $λ$ of spectral lines in hydrogen is given by:

\frac{1}{λ} = R_{H} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})

Where $R_{H} = 1.097 \times 10^{7} m^{- 1}$ is the Rydberg constant. We will calculate $λ$ for each transition.

💡 Teacher's Secret Hint

Remember that $n_{2}$ must always be greater than $n_{1}$ for emission spectra.

Step 2: Calculate Wavelengths for Each Transition○ Expand

A) For $n_{2} = 3$ to $n_{1} = 2$ :

\frac{1}{λ_{A}} = R_{H} (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = R_{H} (\frac{1}{4} - \frac{1}{9}) = R_{H} (\frac{9 - 4}{36}) = \frac{5 R_{H}}{36}

λ_{A} = \frac{36}{5 R_{H}} = \frac{36}{5 \times 1.097 \times 10^{7} m^{- 1}} \approx 6.563 \times 10^{- 7} m = 656.3 nm

B) For $n_{2} = 4$ to $n_{1} = 2$ :

\frac{1}{λ_{B}} = R_{H} (\frac{1}{2^{2}} - \frac{1}{4^{2}}) = R_{H} (\frac{1}{4} - \frac{1}{16}) = R_{H} (\frac{4 - 1}{16}) = \frac{3 R_{H}}{16}

λ_{B} = \frac{16}{3 R_{H}} = \frac{16}{3 \times 1.097 \times 10^{7} m^{- 1}} \approx 4.861 \times 10^{- 7} m = 486.1 nm

C) For $n_{2} = 5$ to $n_{1} = 2$ :

\frac{1}{λ_{C}} = R_{H} (\frac{1}{2^{2}} - \frac{1}{5^{2}}) = R_{H} (\frac{1}{4} - \frac{1}{25}) = R_{H} (\frac{25 - 4}{100}) = \frac{21 R_{H}}{100}

λ_{C} = \frac{100}{21 R_{H}} = \frac{100}{21 \times 1.097 \times 10^{7} m^{- 1}} \approx 4.341 \times 10^{- 7} m = 434.1 nm

D) For $n_{2} = 6$ to $n_{1} = 2$ :

\frac{1}{λ_{D}} = R_{H} (\frac{1}{2^{2}} - \frac{1}{6^{2}}) = R_{H} (\frac{1}{4} - \frac{1}{36}) = R_{H} (\frac{9 - 1}{36}) = \frac{8 R_{H}}{36} = \frac{2 R_{H}}{9}

λ_{D} = \frac{9}{2 R_{H}} = \frac{9}{2 \times 1.097 \times 10^{7} m^{- 1}} \approx 4.102 \times 10^{- 7} m = 410.2 nm

💡 Teacher's Secret Hint

The wavelengths calculated correspond to the visible region of the electromagnetic spectrum, characteristic of the Balmer series.

Step 3: Match Transitions to Wavelengths○ Expand

Based on the calculations:

A) $n_{2} = 3$ to $n_{1} = 2$ corresponds to $656.3 nm$ (List II - III).

B) $n_{2} = 4$ to $n_{1} = 2$ corresponds to $486.1 nm$ (List II - IV).

C) $n_{2} = 5$ to $n_{1} = 2$ corresponds to $434.1 nm$ (List II - II).

D) $n_{2} = 6$ to $n_{1} = 2$ corresponds to $410.2 nm$ (List II - I).

Thus, the correct matching is A-III, B-IV, C-II, D-I. This matches option (1).

💡 Teacher's Secret Hint

As $n_{2}$ increases, the energy difference between levels decreases, leading to shorter wavelengths (higher frequencies).

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Physics Question 2 – NEET-UG 2024

Clue 1 of 3: Hydrogen Spectrum

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