Question

If the energy of spectral line emitted for the transition n = 6 to n = 3 of Li²⁺ is equal to transition in H atom corresponding to which of the following?
1. n = 2 to n = 1
2. n = 3 to n = 2
3. n = 4 to n = 3
4. n = 4 to n = 1

Answer 1

Correct Answer - Option 1 : n = 2 to n = 1

Concept:

Spectrum is the impression produced on a photographic film when the radiation (s) of a particular wavelength (s) is (are) analysed through a prism or diffraction grating.

Types of spectrum           

(1) Emission spectrum:

The spectrum produced by the emitted radiation is known as an emission spectrum. This spectrum corresponds to the radiation emitted (energy evolved) when an excited electron returns back to the ground state.           

(i) Continuous spectrum: When sunlight is passed through a prism, it gets dispersed into continuous bands of different colours. If the light of an incandescent object resolved through prism or spectroscope, it also gives a continuous spectrum of colours.           

(ii) Line spectrum: If the radiation obtained by the excitation of a substance is analysed with help of a spectroscope a series of thin bright lines of specific colours are obtained. There is dark space in between two consecutive lines. This type of spectrum is called a line spectrum or atomic spectrum.  

(2) Absorption spectrum:

The spectrum produced by the absorbed radiations is called absorption spectrum.

The hydrogen spectrum is an example of line emission spectrum or atomic emission spectrum.

The Energy of the various spectral lines (n₂ → n₁) is given by 

\(E = hcR{Z^2}\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right]\)

Calculation:

Given spectral line emitted for the transition n = 6 to n = 3 of Li²⁺ is equal to transition in H atom;

Let the transition in H atom be from n₂ → n₁;

For Li²⁺, Energy of the transition is given by 

\(E = hcR \times {3^2}\left[ {\frac{1}{{{3^2}}} - \frac{1}{{{6^2}}}} \right] = hcR\left[ {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right]\)

Now for the Transition in H atom,

\(E = hcR \times {1^2}\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right] = hcR\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right]\)

Equating both the energies,

⇒ n₁ = 1, n₂ = 2