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

##### 1 Answer
Jun 15, 2016

$4.44 \cdot {10}^{-} 19 J$

#### 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:

$E = h \cdot f$

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:

$f = \frac{c}{\lambda}$

Since the speed of light, $c$, is relatively constant ($3.00 \cdot {10}^{8} \frac{m}{s}$) (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:

$E = \frac{h \cdot c}{\lambda}$
$E = \frac{\left(6.63 \cdot {10}^{-} 34 J \cdot s\right) \cdot \left(3.00 \cdot {10}^{8} \cdot \frac{m}{s}\right)}{448 \cdot {10}^{-} 9 \cdot m}$
$E = 4.44 \cdot {10}^{-} 19 J$