JEE MAIN - Physics (2021 - 31st August Evening Shift - No. 5)
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)
1.45 $$\times$$ 1016 MHz
0.19 $$\times$$ 1015 MHz
1.45 $$\times$$ 109 MHz
9.0 $$\times$$ 1027 MHz
Explanation
For every large distance P.E. = 0
& total energy = 2.6 + 0 = 2.6 eV
Finally in first excited state of H atom total energy = $$-$$3.4 eV
Loss in total energy = 2.6 $$-$$ ($$-$$3.4) = 6 eV
It is emitted as photon
$$\lambda = {{1240} \over 6} = 206$$ nm
$$f = {{3 \times {{10}^8}} \over {206 \times {{10}^{ - 9}}}}$$ = 1.45 $$\times$$ 1015 Hz
= 1.45 $$\times$$ 109 Hz
& total energy = 2.6 + 0 = 2.6 eV
Finally in first excited state of H atom total energy = $$-$$3.4 eV
Loss in total energy = 2.6 $$-$$ ($$-$$3.4) = 6 eV
It is emitted as photon
$$\lambda = {{1240} \over 6} = 206$$ nm
$$f = {{3 \times {{10}^8}} \over {206 \times {{10}^{ - 9}}}}$$ = 1.45 $$\times$$ 1015 Hz
= 1.45 $$\times$$ 109 Hz
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