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.

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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
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