Question

a cook uses a microwave oven to heat a meal. the wavelength of the radiation is 1.20 cm. what is the energy of one photon of this microwave radiation? how many photons required to make popcorn in a 1200 w microwave (180 s time)?

Explanation:

Step1: Recall photon energy formula

The energy of a single photon is given by \( E = h
u \), and since \( c=\lambda
u \) (where \( c \) is the speed of light, \( \lambda \) is wavelength, \(
u \) is frequency), we can substitute \(
u=\frac{c}{\lambda} \) into the photon energy formula to get \( E = \frac{hc}{\lambda} \). Here, \( h = 6.626\times 10^{-34}\ \text{J·s} \), \( c = 3.0\times 10^{8}\ \text{m/s} \), and \( \lambda = 1.20\ \text{cm}=0.0120\ \text{m} \).

Calculate the energy of one photon:
\( E=\frac{(6.626\times 10^{-34}\ \text{J·s})(3.0\times 10^{8}\ \text{m/s})}{0.0120\ \text{m}} \)
\( E=\frac{1.9878\times 10^{-25}\ \text{J·m}}{0.0120\ \text{m}} \)
\( E\approx 1.6565\times 10^{-23}\ \text{J/photon} \)

Step2: Calculate total energy from microwave

Power \( P = 1200\ \text{W}=1200\ \text{J/s} \), time \( t = 180\ \text{s} \). The total energy \( E_{\text{total}} \) is given by \( E_{\text{total}}=P\times t \).

\( E_{\text{total}} = 1200\ \text{J/s}\times 180\ \text{s}=216000\ \text{J} \)

Step3: Find number of photons

The number of photons \( n \) is the total energy divided by the energy per photon: \( n=\frac{E_{\text{total}}}{E} \).

\( n=\frac{216000\ \text{J}}{1.6565\times 10^{-23}\ \text{J/photon}} \approx 1.304\times 10^{28}\ \text{photons} \)

Answer:

The energy of one photon is approximately \( 1.66\times 10^{-23}\ \text{J} \), and the number of photons is approximately \( 1.30\times 10^{28} \) (or more precisely \( 1.304\times 10^{28} \)). For the energy of one photon: \( \boldsymbol{1.66\times 10^{-23}\ \text{J}} \); for the number of photons: \( \boldsymbol{1.30\times 10^{28}} \) (or detailed value as calculated).