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A free electron of 2.6 eV2.6 eV energy collides with a H+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=6.6×10−34 J s)ℎ=6.6×10-34 J sYou Marked | Incorrect AnswerA

Question

A free electron of 2.6 eV2.6 eV energy collides with a H+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=6.6×10−34 J s)ℎ=6.6×10-34 J sYou Marked | Incorrect AnswerA

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Solution

The energy of a free electron is given as 2.6 eV.

1 eV = 1.6 x 10^-19 Joules

So, 2.6 eV = 2.6 * 1.6 x 10^-19 = 4.16 x 10^-19 Joules

The energy of the first excited state of a hydrogen atom is -3.4 eV.

So, -3.4 eV = -3.4 * 1.6 x 10^-19 = -5.44 x 10^-19 Joules

The energy of the photon released is the difference between the energy of the free electron and the energy of the first excited state of the hydrogen atom.

So, Energy of photon = 4.16 x 10^-19 - (-5.44 x 10^-19) = 9.6 x 10^-19 Joules

The frequency of the photon can be calculated using the formula E = h * f, where E is the energy of the photon, h is Planck's constant, and f is the frequency of the photon.

So, f = E / h = 9.6 x 10^-19 / 6.6 x 10^-34 = 1.45 x 10^15 Hz

Therefore, the frequency of the emitted photon is 1.45 x 10^15 Hz.

This problem has been solved

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