A particular hydrogen-like ion emits the radiation of frequency 3 × 10¹⁵ Hz when it makes transition from n=2 to n=1. The frequency of radiation emitted in transition from n=3 to n=1 is x/9 × 10¹⁵ Hz, when x= ________ .

Solution

Physics sir · 3 steps

Take the ratio of the two transitions: Z and the Rydberg constant cancel, so only the 1/n² terms matter.

1. Frequency formula. For a hydrogen-like ion, the emitted frequency is ν = RcZ²(1/n_f²-1/nᵢ²). The ion is the same in both transitions, so RcZ² is a common constant and cancels in a ratio.

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

2. Put in numbers. ν_(3→1) = 32/27 × 3 × 10¹⁵ Hz = 32/9 × 10¹⁵ Hz.
3. So. Matching with x/9 × 10¹⁵ Hz gives x = 32.

The common slip: 3.56. Reporting the frequency coefficient 32/9 ≈ 3.56 gives 3.56, but the blank asks for x, the numerator over 9, which is 32.

The idea to keep. When two transitions belong to the same atom or ion, divide them. Z and R drop out, and you only compare (1/n_f²-1/nᵢ²) for each jump.

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