JEE Main 2020 · 8 January · morning · Single correct · +4 / −1
For the Balmer series in the spectrum of H atom,
ν̅ = R_H{ 1/n₁² - 1/n₂² }, the correct statements among (I) to (IV) are :
(I) As wavelength decreases, the lines in the series converge
(II) The integer n₁ is equal to 2
(III) The lines of longest wavelength corresponds to n₂ = 3
(IV) The ionization energy of hydrogen can be calculated from wave number of these lines
- (a) (II), (III), (IV)
- (b) (I), (II), (III)
- (c) (I), (III), (IV)
- (d) (I), (II), (IV)
Answer and solution
Answer (b) (I), (II), (III)
Solution
Chemistry ma’am · 2 steps
In Balmer every line falls to n = 2, not n = 1, so its series limit is not hydrogen's ionization energy.
- Fix the series. In the Balmer series every line ends on the second orbit, so n₁ = 2 and n₂ = 3, 4, 5, …
ν̅ = R_H(1/2² - 1/n₂²)
Statement Check Verdict (I) Lines converge As n₂ grows, 1/n₂² changes less and less, so the lines crowd together near the series limit at about 365 nm, the short-wavelength end. Correct (II) n₁ = 2 This is what defines the Balmer series. Correct (III) Longest wavelength at n₂ = 3 The smallest gap, 3 → 2, gives the smallest ν̅ and so the longest λ, about 656 nm. Correct (IV) Ionization energy The Balmer limit (n₂ → ∞) gives R_H/4, which is the energy to remove the electron from n = 2, not from the ground state n = 1. Incorrect - So. Statements (I), (II) and (III) are correct, which is option (b).
The idea to keep. Ionization energy of hydrogen means removing the electron from n = 1. Only the Lyman series limit gives it directly. Any other series limit gives the energy from that series' lower level.
Answer from the NTA final answer key. Solution by PrepWiser.