The vapour of Hg absorb some electrons accelerated by a potential difference of 4.5 volts from rest as a result of which light is emitted. If the full energy of a single incident electron is supposed to be converted into light emitted by electron in a single Hg atom, find the wave number 1 λ of the light

Question

The vapour of Hg absorb some electrons accelerated by a potential difference of 4.5 volts from rest as a result of which light is emitted. If the full energy of a single incident electron is supposed to be converted into light emitted by electron in a single Hg atom, find the wave number 1 λ of the light, 3.63 × 10 6 m - 1, 5.93 × 10 6 m - 1, 5.93 × 10 6 c m - 1, 5.62 × 10 6 m - 1

MARKS App · Accepted Answer

Potential difference V = 4.5 eV ∴ Potential energy = 4.5 eV energy absorbed and this energy is emitted by electron of Hg atom. ∴ E = h c λ =4.5 eV = 4.5 × 1.6 × 10 - 19 J ∴ 1 λ = 4.5 × 1.6 × 10 - 19 6.6 × 10 - 34 × 3.8 × 10 8 m - 1 ≈ 3.63 × 10 6 m - 1

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