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.

(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\,=\,6.6474\times 10^14/s

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

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