Question

An AM radio station broadcasts at 1030 kHz , and its FM partner broadcasts at 98.5 MHz .

partA;Calculate the energy of the photons emitted by the AM radio station.

partB;Calculate the energy of the photons emitted by the FM radio station.

partC;Compare the energy of the photons emitted by the AM radio station with the energy of the photons emitted by the FM radio station.

Answer 1

A. The energy emitted by the AM radio ,E = 6.82478.10⁻²⁸ J

B. The energy emitted by the FM radio .E = 6.5266 .10⁻²⁶J

C. ratio  the energy of AM radio : FM radio= 1.0457. 10⁻²

Further explanation

The photoelectric effect is an electron coming out of a metal because of electromagnetic radiation

One type of electromagnetic radiation is light

Electrons can come out of metal because they absorb electromagnetic energy radiated on metals. There is also kinetic energy released from metal, which is according to the equation:

$\large{\boxed{\bold{E=hf-hfo}}}$

fo = the threshold frequency of electromagnetic waves

Radiation energy is absorbed by photons

The energy in one photon can be formulated as

$\rm E=h\times f$

Where

E = energy of light, J

h = Planck's constant (6,626.10⁻³⁴ Js)

f = Frequency of electromagnetic waves, Hz

f = c / λ

c = speed of light

= 3.10⁸

λ = wavelength

• A. An AM radio station broadcasts at 1030 kHz

E = 6,626.10⁻³⁴ Js x 1,030 .10⁶ Hz

E = 6,82478.10⁻²⁸ J

• B. FM broadcasts at 98.5 MHz.

E = 6,626.10⁻³⁴ Js x 98.5 .10⁶ Hz

E = 6.5266 .10⁻²⁶ J

• C. ratio AM radio and FM radio

ratio = 6,82478.10⁻²⁸ J: 6,5266 .10⁻²⁶ J

ratio = 1.0457. 10⁻²

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Answer 2

The energy of the photons emitted by the AM radio station is 6.825 x 10⁻²⁵ J.

The energy of the photons emitted by the FM radio station is 6.53 x 10⁻²⁶ J.

The energy of the photons emitted by the AM radio station is greater than the energy of the photons emitted by the FM radio station.

The given parameters;

• frequency of the AM radio station, f = 1030 kHz
• frequency of the FM radio station, f = 98.5 MHz

The energy of the photons emitted by the AM radio station is calculated as follows;

E = hf

where;

• h is Planck's constant

E = (6.626 x 10⁻³⁴)(1030 x 10³)

E = 6.825 x 10⁻²⁵ J.

The energy of the photons emitted by the FM radio station is calculated as follows;

E = hf

E =  (6.626 x 10⁻³⁴)(98.5 x 10⁶)

E = 6.53 x 10⁻²⁶ J.

Thus, the energy of the photons emitted by the AM radio station is greater than the energy of the photons emitted by the FM radio station.

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