Questions: Microwave ovens emit microwave energy with a wavelength of 12.0 cm. What is the energy of exactly one photon of this microwave radiation? E =

Solution Steps

We are given the wavelength of the microwave radiation, which is \( \lambda = 12.0 \, \text{cm} \). To use this in calculations, we need to convert it to meters:
\[ \lambda = 12.0 \, \text{cm} = 0.12 \, \text{m} \]

The energy of a photon is related to its wavelength by the equation:
\[ E = \frac{hc}{\lambda} \]
where:

• \( E \) is the energy of the photon,
• \( h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \) is Planck's constant,
• \( c = 3.00 \times 10^8 \, \text{m/s} \) is the speed of light,
• \( \lambda \) is the wavelength of the radiation.

Substitute the known values into the equation:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{J} \cdot \text{s})(3.00 \times 10^8 \, \text{m/s})}{0.12 \, \text{m}} \]

Calculate the energy:
\[ E = \frac{1.9878 \times 10^{-25} \, \text{J} \cdot \text{m/s}}{0.12 \, \text{m}} \]
\[ E = 1.6565 \times 10^{-24} \, \text{J} \]

Final Answer