A free electron of 2.6 eV 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 $\times$ 10$-$34 Js)
Solution
For every large distance P.E. = 0<br><br>& total energy = 2.6 + 0 = 2.6 eV<br><br>Finally in first excited state of H atom total energy = $-$3.4 eV<br><br>Loss in total energy = 2.6 $-$ ($-$3.4) = 6 eV<br><br>It is emitted as photon<br><br>$\lambda = {{1240} \over 6} = 206$ nm<br><br>$f = {{3 \times {{10}^8}} \over {206 \times {{10}^{ - 9}}}}$ = 1.45 $\times$ 10<sup>15</sup> Hz<br><br>= 1.45 $\times$ 10<sup>9</sup> Hz
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: Photoelectric Effect
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