The energy levels of an atom is shown in figure.

Which one of these transitions will result in the emission of a photon of wavelength 124.1 nm?
Given ($$h = 6.62 \times 10^{-34}$$ J s)

Solution :

Energy of photon emitted is equal to the difference between the energy levels.

Using,

E = hc/λ

Given :

h = 6.62 × 10⁻³⁴ J s

c = 3 × 10⁸ m/s

λ = 124.1 nm = 124.1 × 10⁻⁹ m

Substituting,

E = (6.62 × 10⁻³⁴ × 3 × 10⁸)/(124.1 × 10⁻⁹)

E ≈ 1.6 × 10⁻¹⁸ J

Now,

1 eV = 1.6 × 10⁻¹⁹ J

Therefore,

E = (1.6 × 10⁻¹⁸)/(1.6 × 10⁻¹⁹)

= 10 eV

From the figure, transition from 0 eV to −10 eV has energy difference of 10 eV.

Hence, transition D will emit photon of wavelength 124.1 nm.

Final Answer :

D