Question

4. Calculate the energy (E) and wavelength (1) of a photon of light with a frequency of6.165 x 10^14 Hz.

Answer 1

A photon of light with frequency of 6.165 x 10^14 Hz has a wavelength of 486.6 nm and energy of 4.084x10^-19 J

The energy of a photon is defined as

`E\text{ = h}* f`

where h is the Planck's constant and f is the frequency.

h is a constant of value h = 6.62607004 × 10^-34 m2 kg / s

and f was given in the question, f = 6.165 x 10^14 Hz (keep in mind that 1 Hz = 1/s)

Then we calculate:

`E=6.165*10^(14)Hz\text{ }*6.626*10^(-34)m2kg/s=4.084\text{ }*10^(-19)m^2kg/s^2`

Since the unit Joules corresponds to (m^2 kg / s^2), we have that E = 4.084x10^-19 J, and we may calculate the wavelength of the photon considering that:

`\lambda=(c)/(f)`

where λ is the wavelength, c is the velocity of light and f is the frequency.

`\lambda=(300000000(m)/(s))/(6.165*10^(14)s^(-1))=4.866*10^(-7)m=486.6\text{ }*10^(-9)m=486.6\text{ nm}`