A particular hydrogen like ion emits radiation of frequency 2.92 × 10¹⁵ Hz when it makes transition from n = 3 to n = 1. The frequency in Hz of radiation emitted in transition from n = 2 to n = 1 will be :

Solution

Physics sir · 3 steps

You do not need Z here: take the ratio of the two frequencies and Z cancels.

1. Frequency of a line. For a hydrogen-like ion, the emitted frequency is ν = RcZ²(1/n₁²-1/n₂²). Both transitions are in the same ion, so RcZ² is the same and drops out of the ratio.

   ν_(2→1)/ν_(3→1) = (1/1²-1/2²)/(1/1²-1/3²) = (3/4)/(8/9) = 27/32

2. Put in numbers. ν_(2→1) = 27/32 × 2.92 × 10¹⁵ = 2.46 × 10¹⁵ Hz. It is smaller, as it should be, because the 2 → 1 jump is smaller than the 3 → 1 jump.
3. So. The frequency for n = 2 → 1 is 2.46 × 10¹⁵ Hz, option (d).

The idea to keep. For two lines of the same ion, compare the brackets (1/n₁²-1/n₂²). Z and the Rydberg constant cancel, so never waste time finding them.

Answer from the NTA final answer key. Solution by PrepWiser.