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]
- 1.3 x 10-15 Hz
- 3.1 x 1052 Hz
- 3.2 x 10-53 Hz
-
8.0 x 1014 Hz
The correct answer is: D
Explanation
E = hff = (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