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A free electron of energy 1. 4 e V colli...

A free electron of energy 1. 4 e V collides with a `H^(+)` ion. As a result of collision a hydrogen atom in the ground state is formed and a photon is released. What is the wavelength of the emitted photon? In which part of the electromagnetic spectrum does this wavelength lie?
[Given, ionisation potential of hydrogen =13.6 eV]

Text Solution

Verified by Experts

In ground state, the electron will have -13.6 eV energy.
So, `1.4 eV = - 13.6 eV + (hv)/lambda`
`(hc)/lambda = 1.4 + 13.6 = 15eV`
`lambda = (hc)/(15) = (6.6 xx 10^(-34) xx3xx10^(8))/(15 xx 1.6 xx 10^(-19))`
` = 0.825 xx 10^(-7) = 825 Å`
It lies in UV-region of electromagnetic spectrum.
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