Question

Find the quantum ( photon) energy associated with a 650.kHz AM radio signal.

Answer 1

The energy in one photon is:

$$E = h\nu$$

where $h = 6.63\times 10^{-34}J\cdot s$ is the Planck's constant, and $\nu = 650kHz = 6.5\times 10^5Hz$ is the frequency of the light. Thus, obtain:

$$E = 6.63\times 10^{-34}J\cdot s\cdot 6.5\times 10^5Hz = 1.02\times 10^{-29}J$$

Answer. $1.02\times 10^{-29}J$.