The energy associated with the emitted photon when a mercury atom changes from one state to another is 3.3 ev. calculate the frequency of the photon.
[e = 1.6 x 10-19c; h = 6.6 x 10-13js]
[e = 1.6 x 10-19c; h = 6.6 x 10-13js]
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Correct Answer: Option D
Explanation:
E = hf
f = (E)/h
∴ f = (3.3 x 1.6 x 10-19)/6.6 x 10-34
= 0.8 x 1015 = 8.0 x 1014 Hz
E = hf
f = (E)/h
∴ f = (3.3 x 1.6 x 10-19)/6.6 x 10-34
= 0.8 x 1015 = 8.0 x 1014 Hz