Chemistry Q46 – NEET-UG 2025

Home Chemistry Structure of AtomQuestion #608

The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes

n = 2 \to n = 3

and

n = 4 \to n = 6

transitions, respectively, is

Clue 1 of 3: Atomic Spectra

The energy of light absorbed or emitted during electron transitions in a hydrogen atom is quantized and can be related to its wavelength.

🧠 Full Solution Path

Step 1: Apply Rydberg Formula for First Transition✦ Active

For a hydrogen atom ( $Z = 1$ ), the Rydberg formula for the reciprocal of the wavelength is given by $\frac{1}{λ} = R_{H} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$ . For the first transition $n = 2 \to n = 3$ , we have $n_{1} = 2$ and $n_{2} = 3$ .

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

Thus, $λ_{1} = \frac{36}{5 R_{H}}$ .

Step 2: Apply Rydberg Formula for Second Transition○ Expand

For the second transition $n = 4 \to n = 6$ , we have $n_{1} = 4$ and $n_{2} = 6$ .

\frac{1}{λ_{2}} = R_{H} (\frac{1}{4^{2}} - \frac{1}{6^{2}}) = R_{H} (\frac{1}{16} - \frac{1}{36})

To subtract the fractions, find a common denominator, which is 144.

\frac{1}{λ_{2}} = R_{H} (\frac{9}{144} - \frac{4}{144}) = R_{H} (\frac{5}{144})

Thus, $λ_{2} = \frac{144}{5 R_{H}}$ .

💡 Teacher's Secret Hint

Remember to find the least common multiple for the denominators when subtracting fractions.

Step 3: Calculate the Ratio of Wavelengths○ Expand

Now, calculate the ratio of the wavelengths $\frac{λ_{1}}{λ_{2}}$ .

\frac{λ_{1}}{λ_{2}} = \frac{\frac{36}{5 R_{H}}}{\frac{144}{5 R_{H}}} = \frac{36}{144}

Simplify the fraction:

\frac{36}{144} = \frac{1}{4}

💡 Teacher's Secret Hint

The Rydberg constant $R_{H}$ cancels out in the ratio, simplifying the calculation.

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Chemistry Question 46 – NEET-UG 2025

Clue 1 of 3: Atomic Spectra

🧠 Full Solution Path