Question

What is the energy in joules and eV of a photon in a radio wave from an AM station that has a 1530-kHz broadcast frequency?

a. (1.30 × 10⁻¹⁵) J, (8.13 × 10⁻¹⁹) eV
b. (8.63 × 10⁻²⁸) J, (5.39 × 10⁻²¹) eV
c. (1.99 × 10⁻²⁵) J, (1.24 × 10⁻¹⁷) eV
d. (1.32 × 10⁻³⁴) J, (8.26 × 10⁻²⁸) eV

Answer 1

The energy of a photon from an AM radio station at a frequency of 1530 kHz is 1.014 × 10−15 J, which can also be converted to 6.33 × 10−19 eV.

Step-by-step explanation:

To find the energy of a photon emitted by an AM radio station with a frequency of 1530 kHz, we can use the formula E = hf, where h is Planck's constant (6.626 × 10−34 J·s), and f is the frequency of the photon. First, convert the frequency from kHz to Hz by multiplying by 103, getting 1530 × 103 Hz or 1.530 × 106 Hz. Then, multiply by Planck's constant to get the energy in joules (J). To convert the energy into electronvolts (eV), use the conversion factor of 1 eV = 1.602 × 10−19 J.

Using this method, the energy in joules (J) is calculated as:
E = (6.626 × 10−34 J·s) (1.530 × 106 Hz) = 1.014 × 10−15 J.

To convert this energy to eV:
E = 1.014 × 10−15 J × (1 eV / 1.602 × 10−19 J) = 6.33 × 10−19 eV.