A cell phone emits signs in approximately 750 Mhz. Which is the wavelength and energy of this radiation

To determine the wavelength and energy of the radiation emitted by a cell phone operating at approximately 750 MHz, we'll need to apply some basic equations.

First, we'll start with the equation:

Speed of Light (c) = Wavelength (λ) * Frequency (f).

The speed of light is a constant value approximately equal to 3 x 10^8 meters per second (m/s).

Given that the frequency is 750 MHz, we need to convert it into units of Hz:

750 MHz = 750 x 10^6 Hz.

Now we can rearrange the equation to solve for the wavelength:

λ = c / f.

Substituting the values:

λ = (3 x 10^8 m/s) / (750 x 10^6 Hz).

λ = 0.4 meters.

Therefore, the wavelength of the radiation emitted by the cell phone is approximately 0.4 meters (or 40 centimeters).

To calculate the energy of the radiation, we'll use the equation:

Energy (E) = Planck's Constant (h) * Frequency (f),

where Planck's Constant (h) is approximately 6.63 x 10^-34 Joule-seconds (J·s).

Substituting the values:

E = (6.63 x 10^-34 J·s) * (750 x 10^6 Hz).

E = 4.9725 x 10^-26 Joules.

Therefore, the energy of the radiation emitted by the cell phone is approximately 4.9725 x 10^-26 Joules.