A free electron of 2.6 e V energy collides with a H + ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photo

Question

A free electron of 2.6 e V energy collides with a H + ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon. ( h = 6.6 × 1 0 − 34 J s )

TestHub · Accepted Answer

By Energy conservation ( K . E . ) e ​ = ( T . E . ) H + ​ + E P h o t o n ​ 2.6 = 4 − 13.6 ​ + h f h f = 6 e V f = 6.626 × 1 0 − 34 6 × 1.6 × 1 0 − 19 ​ f = 1.45 × 1 0 15 Hz = 1.45 × 1 0 9 MHz

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