Indium compounds give a blue-violet flame test


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

 

This question can be found in General Chemistry, 9th edition, chapter 7, question 7.46

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