Class 11 Chemistry Chapter 2: Structure of Atom case study question with answers
One case-based question of the kind CBSE-pattern school exams ask: a passage, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.
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Answer: (c) Balmer series
The Balmer series (ending at n = 2) is the only series of hydrogen lines in the visible region.
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Answer: (a) n = 3 to n = 2
In a series the longest wavelength comes from the smallest jump, n₂ = n₁ + 1: 3 to 2. 6 to 2 gives the violet line; 2 to 1 is ultraviolet and 4 to 3 infrared.
Answer and marking scheme
From the line: ΔE = hc/λ = (6.626 × 10⁻³⁴ × 3.0 × 10⁸)/(486 × 10⁻⁹) = 4.09 × 10⁻¹⁹ J. From Bohr: ΔE = 2.18 × 10⁻¹⁸ (1/2² − 1/4²) = 2.18 × 10⁻¹⁸ × 3/16 = 4.09 × 10⁻¹⁹ J. The two agree.
- ΔE = hc/λ = 4.09 × 10⁻¹⁹ J (1)
- Bohr ΔE with n = 4 and 2 = 4.09 × 10⁻¹⁹ J (1)
Answer and marking scheme
The electron's energy in a hydrogen atom is quantised; a jump between two fixed levels releases a photon of one definite energy and so one wavelength, giving a line. When white light passes through hydrogen, the atoms absorb only the wavelengths that match jumps between their levels: an absorption spectrum, a continuous band crossed by dark lines, like a photographic negative of the emission spectrum (the dark lines lie at wavelengths where hydrogen can also emit).
- quantised levels: each transition gives one wavelength (1)
- absorption spectrum: dark lines on a continuous background, the negative of the emission spectrum (1)
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