3 free questions from this chapter (the full set has 30). Marking: +4 for correct · −1 for incorrect · 0 for unattempted.
Question 1Level 3 · AdvancedNumerical answer
A large sample of hydrogen atoms has all its electrons in the n=6 level. As the atoms return to the ground state by every possible route, how many distinct spectral lines lie OUTSIDE the visible region (take visible as 400 nm to 700 nm)? Use RH=1.097×107 m−1.
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Answer: 11
Key insight: all visible lines of hydrogen are Balmer lines, so count Balmer lines available from n=6 and subtract from the total. Total distinct lines from n=6 in a sample =26×5=15. Balmer lines available: 3→2 (656 nm), 4→2 (486 nm), 5→2 (434 nm), 6→2: λ1=RH(41−361) gives λ≈410 nm, still visible. So 4 lines are visible. Lyman lines are UV and Paschen/Brackett/Pfund lines are IR. Lines outside visible =15−4=11. Trap: assuming 6→2 falls in the UV (the first Balmer line in the UV is 7→2 at about 397 nm). Exam pattern: JEE Main – number of spectral lines from a given excited levelQuestion 2Level 3 · AdvancedNumerical answer
A SINGLE hydrogen atom in the n=6 level returns to the ground state by emitting photons one after another. It is known that one of the photons emitted is the Hβ line of the Balmer series (n=4→n=2). What is the maximum number of photons this atom could have emitted?
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Answer: 4
Key insight: a single atom follows one cascade, and the forced 4→2 step fixes part of the route. The atom must reach n=4 from n=6: at most 2 photons (6→5, 5→4). Then 4→2 is one photon. From n=2 to n=1 only one photon is possible. Maximum =2+1+1=4. Trap 1: answering 15 (that is the sample count). Trap 2: answering 5 (n−1 for an unconstrained single atom), which would need the route 4→3→2, excluding Hβ. Exam pattern: JEE Main – maximum photons emitted by a single atomQuestion 3Level 3 · AdvancedSingle correct
Which transition in the Li2+ ion emits radiation of exactly the same wavelength as the Hα line (n=3→n=2) of hydrogen?
An=3→n=2
Bn=6→n=3
Cn=9→n=6
Dn=9→n=4
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Answer: C
Key insight: λ1=RHZ2(n121−n221), and multiplying both n values by Z cancels the Z2. For Z=3: n1=3×2=6, n2=3×3=9. Check: 9(361−811)=9⋅3245=365=41−91. Option B (6→3) gives 9(91−361)=43, the Lyman-α value. Option A ignores Z2. Option D wrongly scales by Z2 for one level only. Exam pattern: JEE Main – identical wavelength lines in H and hydrogen-like ions