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MCQEasyJEE 2026Bohr Model & Hydrogen Spectrum

JEE Chemistry 2026 Question with Solution

The energy of first (lowest) Balmer line of H atom is xx J. The energy (in J) of second Balmer line of H atom is:

  • A

    x1.35\dfrac{x}{1.35}

  • B

    x2x^2

  • C

    1.35x1.35x

  • D

    2x2x

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The energy of the first Balmer line is xx.

Find: The energy of the second Balmer line.

Balmer series corresponds to transitions ending at n=2n=2.

First (lowest) Balmer line:

n=3→2n=3 \rightarrow 2

Second Balmer line:

n=4→2n=4 \rightarrow 2

Use the energy expression for hydrogen spectral lines:

E=13.6(1n12−1n22) eVE = 13.6\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\text{ eV}

For the first Balmer line:

E1=13.6(122−132)=13.6(536)E_1 = 13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right) = 13.6\left(\frac{5}{36}\right)

For the second Balmer line:

E2=13.6(122−142)=13.6(316)E_2 = 13.6\left(\frac{1}{2^2}-\frac{1}{4^2}\right) = 13.6\left(\frac{3}{16}\right)

Now find the ratio:

E2E1=316536=2720≈1.35\frac{E_2}{E_1} = \frac{\frac{3}{16}}{\frac{5}{36}} = \frac{27}{20} \approx 1.35

Therefore,

E2=1.35 E1=1.35xE_2 = 1.35\,E_1 = 1.35x

So, the correct option is C.

Energy Ratio View

Given: First Balmer line energy is xx.

Find: Second Balmer line energy in terms of xx.

In the Balmer series, all transitions terminate at n=2n=2. The first two Balmer lines are 3→23 \rightarrow 2 and 4→24 \rightarrow 2.

Since emitted photon energy equals the difference between the two energy levels, compare the two transition energies directly:

E3→2∝(122−132)=536E_{3 \rightarrow 2} \propto \left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\frac{5}{36} E4→2∝(122−142)=316E_{4 \rightarrow 2} \propto \left(\frac{1}{2^2}-\frac{1}{4^2}\right)=\frac{3}{16}

Hence,

E4→2E3→2=316536=2720=1.35\frac{E_{4 \rightarrow 2}}{E_{3 \rightarrow 2}}=\frac{\frac{3}{16}}{\frac{5}{36}}=\frac{27}{20}=1.35

Therefore the second Balmer line has energy 1.35x1.35x.

Common mistakes

  • Confusing the Balmer series with transitions starting from n=2n=2 instead of ending at n=2n=2 is incorrect. In the Balmer series, the final level is fixed at n=2n=2. Always identify the correct terminal level before writing the transition.

  • Using the first Balmer line as n=2→1n=2 \rightarrow 1 is wrong because that belongs to the Lyman series. The first Balmer line is n=3→2n=3 \rightarrow 2. Check the series definition before substituting values.

  • Assuming the second Balmer line has double the energy of the first is incorrect because hydrogen energy gaps are not equally spaced. Use the level-difference formula and compare the actual values of the transitions.

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