Question

An FM radio station broadcasts at 98 MHz. what is the energy of each photon in Joule? Use h= 6.6 X10^-34 J*s for Planck constant.

Answer 1

The energy of each photon is
`6.468 * 10^(-26)` Joule.

Explanation:

Consider the provided information.

According to the plank equation:

`E=h\\u`

Where E is the energy of photon, h is the plank constant and
`\\u` is the frequency.

It is given that
`h= 6.6 *10^(-34)` and
`\\u=98MHz`

98Mhz =
`98* 10^6Hz`

Substitute the respective value in plank equation.

`E=6.6* 10^(-34)* 98* 10^6`

`E=6.6* 98* 10^(-34+6)`

`E=646.8 *10^(-28)`

`E=6.468 * 10^(-26)`

Hence, the energy of each photon is
`6.468 * 10^(-26)` Joule.