Question

If '\(\omega\)' is the angular frequency and '\(k\)' is the wave number representing a photon of energy \(E\) and momentum \(P\) through the wave function \[ \psi(x,t)=A\sin(kx-\omega t) \] Which of the following relation is true

• A. \(E=kP\)
• B. \(P=\omega E\)
• C. \(P=\dfrac{\omega}{k}E\)
• D. \(E=\dfrac{\omega}{k}P\)

Hint:

Photon energy is proportional to angular frequency.

Answer 1

The Correct Option is D

Concept:
Photon energy and momentum are related to angular frequency and wave number.
Step 1:
For photon: \[ E=\hbar\omega \]
Step 2:
Momentum is: \[ P=\hbar k \]
Step 3:
Dividing equations: \[ \frac{E}{P}=\frac{\omega}{k} \]
Step 4:
\[ E=\frac{\omega}{k}P \] \[ \boxed{\text{Correct Option = (4)}} \]