Physics JAMB

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]

A. 1.3 x 10-15 Hz
B. 3.1 x 1052 Hz
C. 3.2 x 10-53 Hz
D. 8.0 x 1014 Hz

Correct Answer:

Option D = 8.0 x 1014 Hz

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
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