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.