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MCQMediumJEE 2026Bohr Model & Hydrogen Spectrum

JEE Chemistry 2026 Question with Solution

Given below are two statements:

Statement I: When an electric discharge is passed through gaseous hydrogen, the hydrogen molecules dissociate and the energetically excited hydrogen atoms produce electromagnetic radiation of discrete frequencies.

Statement II: The frequency of second line of Balmer series obtained from He⁺ is equal to that of first line of Lyman series obtained from hydrogen atom.

In the light of the above statements, choose the correct answer from the options given below:

  • A

    Statement I is true but Statement II is false

  • B

    Both Statement I and Statement II are true

  • C

    Statement I is false but Statement II is true

  • D

    Both Statement I and Statement II are false

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Two statements about hydrogen spectrum and hydrogen-like species are to be checked.

Find: Which option correctly identifies the truth values of Statement I and Statement II.

Statement I: When electric discharge is passed through gaseous hydrogen, hydrogen molecules dissociate into atoms. These hydrogen atoms get excited and on returning to lower energy levels emit electromagnetic radiation of discrete frequencies. This is the origin of the line spectrum of hydrogen.

Therefore, Statement I is true.

Statement II: For hydrogen-like species, frequency is given by

ν=cRHZ2(1n12−1n22)\nu = c R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)

For the second line of Balmer series of He⁺:

  • Balmer series means n1=2n_1 = 2
  • Second line means n2=4n_2 = 4
  • For He⁺, Z=2Z = 2

So,

νHe+=cRH(2)2(122−142)\nu_{\mathrm{He}^+} = c R_H (2)^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) νHe+=cRH⋅4(14−116)\nu_{\mathrm{He}^+} = c R_H \cdot 4 \left( \frac{1}{4} - \frac{1}{16} \right) νHe+=cRH⋅4(316)=34cRH\nu_{\mathrm{He}^+} = c R_H \cdot 4 \left( \frac{3}{16} \right) = \frac{3}{4} c R_H

For the first line of Lyman series of hydrogen:

  • Lyman series means n1=1n_1 = 1
  • First line means n2=2n_2 = 2
  • For hydrogen, Z=1Z = 1

So,

νH=cRH(1)2(112−122)\nu_{\mathrm{H}} = c R_H (1)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) νH=cRH(1−14)=34cRH\nu_{\mathrm{H}} = c R_H \left( 1 - \frac{1}{4} \right) = \frac{3}{4} c R_H

Hence,

νHe+=νH\nu_{\mathrm{He}^+} = \nu_{\mathrm{H}}

Therefore, Statement II is true.

So, both Statement I and Statement II are true. The correct option is B.

Series Matching Shortcut

Given: A comparison between the second Balmer line of He⁺ and the first Lyman line of hydrogen.

Find: Whether the two frequencies are equal.

For hydrogen-like species, energy levels vary as Z2n2\dfrac{Z^2}{n^2}. So two transitions have the same frequency when the ratios Zn\dfrac{Z}{n} at initial and final levels match.

For He⁺ second Balmer line, transition is 4→24 \to 2 with Z=2Z = 2:

  • initial ratio: 24=12\dfrac{2}{4} = \dfrac{1}{2}
  • final ratio: 22=1\dfrac{2}{2} = 1

For hydrogen first Lyman line, transition is 2→12 \to 1 with Z=1Z = 1:

  • initial ratio: 12\dfrac{1}{2}
  • final ratio: 11

Both ratios match, so both transitions have the same energy gap and hence the same frequency.

Thus Statement II is true. Statement I is also true because hydrogen line spectrum arises from excitation and de-excitation of hydrogen atoms.

Therefore, the correct option is B.

Common mistakes

  • Assuming gaseous hydrogen directly gives atomic spectral lines without dissociation is incorrect. Molecular hydrogen first dissociates into hydrogen atoms; the observed discrete atomic spectrum comes from electronic transitions in these atoms.

  • Taking the second Balmer line as 3→23 \to 2 is wrong. The first Balmer line is 3→23 \to 2, so the second Balmer line is 4→24 \to 2.

  • Forgetting the Z2Z^2 factor for hydrogen-like ions leads to an incorrect comparison between He⁺ and hydrogen. Always use the full Rydberg expression with the nuclear charge term.

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