Question

The figure shows energy level diagram of hydrogen atom.

(i) Find out the transition which results in the emission of a photon of wavelength 496 nm.

(ii) Which transition corresponds to the emission of radiation of maximum wavelength ? Justify your answer.

Answer 1

(a) For hydrogen atom,

E₁ = -13.6 eV

E₂ = -3.4 eV

E₃ = -1.51 eV

E₄ = -0.85 eV

h = 6.63 x 10^-34 Js;

 c = 3 × 10⁸ ms^-1

Photon Energy = \(\frac {hc}\lambda\)

\(= \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{496 \times 10^{-9} \times 1.6 \times 10^{-19}}eV\)

\(= 2.5 \,eV\)

This equals (nearly) the difference (E₄ - E₂). Hence the required transition is (n = 4) to (n = 2).

[Alternatively: If the candidate calculates by using Rydberg formula, and identifies correctly the required transition, full credit may be given.]

(b) The transition n = 4 to n = 3 corresponds to emission of radiation of maximum wavelength. It is so because this transmission gives out the photon of least energy.

Answer 2

(i) For hydrogen atom,

This equals (nearly) the difference (E₄ -E₂).

Hence the required transition is (n =  4) to (n  = 2)

[Alternatively : If the candidate calculates by using Rydberg formula, and identifies correctly the required transition, full credit may be given.]

(ii)The transition n =  4 to n = 3 corresponds to emission of radiation of maximum wavelength. It is so because this transmission gives out the photon of least energy.