Question

In the sample of H-atoms electrons makes a transition from a higher orbit to ground state. If the maximum number of spectral lines obtained in the spectrum is 15 then what should be the minimum number of H-atoms in the sample to produce these 15 lines.

• A. 6
• B. 15
• C. 10
• D. 9

Answer 1

Answer: (A)

6

Answer 2

To solve this problem, we first need to determine the principal quantum number of the higher orbit from which the electrons are transitioning.

1. Determine the initial principal quantum number (n): The maximum number of spectral lines obtained when electrons transition from a higher orbit 'n' to the ground state (n=1) is given by the formula: $N_{lines} = \frac{n(n-1)}{2}$ We are given that the maximum number of spectral lines is 15. So, $15 = \frac{n(n-1)}{2}$ $30 = n(n-1)$ By inspection, we can find an integer 'n' that satisfies this equation. If $n=6$, then $6 \times (6-1) = 6 \times 5 = 30$. Therefore, the electrons are transitioning from the 6th orbit (n=6) to the ground state.

2. Determine the minimum number of H-atoms: To observe all possible spectral lines (the maximum number of lines) from an initial state 'n' down to the ground state, we need a sample of H-atoms. Each atom, when excited to level 'n', will de-excite, producing one or more photons. To ensure that all possible distinct transitions are observed, we need a sufficient number of atoms.

   Consider the transitions from n=6:

   • Lyman series (to n=1): 6→1, 5→1, 4→1, 3→1, 2→1 (5 lines)
   • Balmer series (to n=2): 6→2, 5→2, 4→2, 3→2 (4 lines)
   • Paschen series (to n=3): 6→3, 5→3, 4→3 (3 lines)
   • Brackett series (to n=4): 6→4, 5→4 (2 lines)
   • Pfund series (to n=5): 6→5 (1 line) Total = 5 + 4 + 3 + 2 + 1 = 15 lines.

   To observe all these 15 distinct lines, we need a minimum number of H-atoms such that each atom, when excited to the highest level (n=6), can contribute to observing the full spectrum. If we have 'n' atoms, each excited to the n-th level, it is generally considered that these 'n' atoms can collectively produce all $n(n-1)/2$ possible transitions. However, a more precise understanding often used in competitive exams is that the minimum number of atoms required to produce all possible lines from an initial state 'n' is 'n'. Each atom can be thought of as undergoing one of the primary transitions from 'n', and then cascading further.

   For example, if n=6, we need 6 atoms:

   • Atom 1: Transitions 6→1 (and potentially other steps if it were a cascade, but the point is it starts at 6)
   • Atom 2: Transitions 6→2
   • Atom 3: Transitions 6→3
   • Atom 4: Transitions 6→4
   • Atom 5: Transitions 6→5
   • Atom 6: This atom is needed to ensure that all intermediate transitions (like 5→4, 4→3, etc., which are not direct from 6) can also be observed. If the first 5 atoms cover the direct transitions from n=6, then the subsequent de-excitations from these atoms will cover the rest of the lines.

   The most common interpretation for this type of problem, especially in JEE/NEET context, is that the minimum number of atoms required to produce all possible spectral lines from an initial state 'n' is 'n'.

   In this case, n=6. Therefore, the minimum number of H-atoms required is 6.