# Indium compounds give a blue-violet flame test 2

Indium compounds give a blue-violet flame test. The atomic emission responsible for this blue-violet color has a wavelength of 451 nm. Obtain the energy of a single photon of this wavelength.

#### Solution:

We will first find the frequency. To do so, we will write the wavelength, 451 nm in meters.

$451\text{nm}\,=\,4.51\times 10^{-7}$ m

Next, we will use this equation to find the frequency.

$v=\dfrac{c}{\lambda}$

(where $v$ is frequency, $c$ is the speed of light, and $\lambda$ is the wavelength)

Speed of light = $2.998\times 10^8 \dfrac{m}{s}$

Substituting the values gives us:

$v\,=\,\dfrac{2.998\times 10^8 \dfrac{m}{s}}{4.51\times 10^-7\text{m}}$

$v\,=\,6.6474\times 10^14$/s

To figure out the energy of a single photon, we will use the following equation:

$E\,=\,hv$

(Where $E$ is photon energy, $h$ is Plank’s Constant, and $v$ is frequency)

Remember that Plank’s Constant = $6.626\times 10^{-34}\text{J}\cdot\text{s}$

Again, substituting the values we have:

$E\,=\,(6.626\times 10^{-34}\text{J}\cdot\text{s})(6.6474\times 10^{14}\text{/s})$

$E\,=\,4.404\times 10^{-19}$ J