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...
Question 1 Report
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]
Answer Details
To calculate the frequency of the photon, we can use the formula E = hf, where E is the energy of the photon, h is Planck's constant, and f is the frequency of the photon. Rearranging the formula to solve for f, we get f = E/h.
Given that the energy associated with the emitted photon when a mercury atom changes from one state to another is 3.3 eV, we first need to convert this to joules. We know that 1 eV is equal to 1.6 x 10^-19 J, so 3.3 eV is equal to 3.3 x 1.6 x 10^-19 J, which is 5.28 x 10^-19 J.
We also know that Planck's constant, h, is 6.6 x 10^-34 J s.
Plugging these values into the formula f = E/h, we get:
f = (5.28 x 10^-19 J)/(6.6 x 10^-34 J s) = 8.0 x 10^14 Hz
Therefore, the frequency of the photon is 8.0 x 10^14 Hz.
Answer: 8.0 x 10^14 Hz
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