The second is defined as the time it takes for 9,192,631,770 wavelengths of a certain transition of the cesium-133 atom to pass a fixed point. What is the frequency of this electromagnetic radiation? What is the wavelength?

#### Solution:

The frequency of this electromagnetic radiation is already given to us in the question. It is simply 9,192,631,770 per second.

To calculate the wavelength, we will use this formula:

\lambda=\dfrac{c}{v}(Where \lambda is the wavelength, c is the speed of light, and v is frequency)

Speed of light = 2.99792458\times 10^8\dfrac{m}{s}

Substituting the values we know and solving for \lambda gives us:

\lambda=\dfrac{2.99792458\times 10^8\dfrac{m}{s}}{9,192,631,770/s}

\lambda=0.0326122557 m

###### This question can be found in General Chemistry, 9th edition, chapter 7, question 7.42

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