Question

What is the energy of a photon emitted with a wavelength of 448 nm?

Answer 1

`4.44 * 10^-19J`

Explanation:

A photon is a particle of light that is released when an electron transitions from a high energy state to a low energy state.

The energy associated with the process is given by the following equation:

`E=h*f`

where `E` is the energy associated with the photon, `h` is Planck's constant (`6.63*10^-34J*s)`, and `f` is the characteristic frequency associated with the photon (expressed in Hertz, or `s^-1`).

With the information given in the problem, it might at first seem as if the problem cannot be solved. However, frequency, `f`, can be related to wavelength, `lambda`, in the following manner:

`f=c/lambda`

Since the speed of light, `c`, is relatively constant (`3.00*10^8m/s`) (this is true for a vacuum, but light slows down in other mediums) and wavelength is known, the second equation can be substituted into the first:

`E=(h*c)/lambda`
`E=((6.63*10^-34J*s)*(3.00*10^8*m/s))/(448*10^-9*m)`
`E=4.44*10^-19J`