Photoelectrons are emitted when 40 nm radiation is

Question

Photoelectrons are emitted when 40 nm radiation is incident on a surface of work function 1.9 eV. These photoelectrons pass through a region containing α – particles. A maximum energy electron combines with an α-particle to form a He+ ion, emitting a single photon in this process. He+ ions thus formed are in their fourth excited state. Find the energies in eV of the photons, lying in the 2 to 4 eV rang, that are likely to be emitted during and after the combination [Take h = 4.14 x 1015 eV.s.]

Navjyot Kalra · Accepted Answer

Hello Student,

Please find the answer to your question

The energy of the incident photon is

E₁ = hc / λ = (4.14 x 10^-15 eVs) (3 x 10⁸ m/s) / (400 x 10^-9m) = 3.1 eV

The maximum kinetic energy of the electrons is E_max = E₁ – W = 3.1 eV – 1.9 eV = 1.2 eV

It is given that,

The fourth excited state implies that the electron enter I the n = 5 state.

In this state its energy is

E₅ = - (13.6 eV)Z² / n² = - (13.6eV)(2)² / 5²

= - 2.18 eV

This energy of the emitted photon in the above combination reaction is

E =E_max + (- E₅) = 1.2 eV + 2.18 eV = 2.4 eV

Note : After the recombination reaction, the electron may undergo transition from a higher level to a lower level thereby emitting photons.

The energies in the electronic levels of He⁺ are

E₄ = (- 13.6 eV)(2²) / 4² = - 3.4 eV

E₃ = ( - 13. 6eV)(2²) / 3² = -6.04 eV

E₂ = (-13.6eV) (2²) / 2² = - 13.6 eV

The possible transitions are

n = 5 ⟶ n = 4

∆E = E₅ – E₄ = [ - 2.18 – ( - 3.4 )] eV = 1.28 eV

n = 5 ⟶ n = 3

∆E = E₅ – E₃ = [ - 2.18 – ( -6.04 )] eV = 3.84 eV

n = 5 ⟶ n = 2

∆E = E₅ – E₂ = [ - 2.18 – ( - 13.6 )] eV = 11.4 eV

n = 4 ⟶ n = 3

∆E = E₄ – E₃ = [ - 3.4 – ( - 6.04 )] eV = 2.64 eV

Thanks
Navjot Kalra
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