PHYSICS
JAMB 2016 - Question 13
Physics 2016 JAMB Past Questions - Question 13: 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. 15 C ; h=6.6 x 10Js
Choose the correct answers from the options given.
A:
B:
C:
D:
Correct Answer
D
Explanation
To calculate the frequency of the photon, we can use the equation:
E = hf
Where:
E is the energy of the photon (in joules),
h is Planck's constant (6.6 x 10^-34 Js),
and f is the frequency of the photon (in hertz).
First, let's convert the energy from electron volts (eV) to joules (J). We know that 1 eV is equal to 1.6 x 10^-19 J.
Given that the energy associated with the emitted photon is 3.3 eV, we can calculate the energy in joules:
E = 3.3 eV * (1.6 x 10^-19 J/eV)
E = 5.28 x 10^-19 J
Now, we can rearrange the equation to solve for the frequency:
f = E / h
f = (5.28 x 10^-19 J) / (6.6 x 10^-34 Js)
f ≈ 8.00 x 10^14 Hz
Therefore, the frequency of the photon is approximately 8.00 x 10^14 Hz.
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