Question

A 640 -nm laser emits a 1 -s pulse in a beam with a diameter of $1.5 \mathrm{mm} .\( The rms electric field of the pulse is \)120 \mathrm{V} / \mathrm{m} .$ How many photons are emitted per second? [Hint: Review Section \(22.6 .]\)

Short answer

Answer: The laser emits approximately \(1.082 \times 10^{17}\) photons per second.

Step-by-step solution

Calculate the intensity of the electromagnetic wave

Recall that the intensity of an electromagnetic wave is given by the formula \(I=\frac{1}{2}\varepsilon_0 c E_{rms}^2,\) where \(\varepsilon_0\) is the vacuum permittivity \((8.85 \times 10^{-12} \ \mathrm{C^2/N \cdot m^2}),\) \(c\) is the speed of light \((3 \times 10^8 \ \mathrm{m/s}),\) and \(E_{rms}\) is the rms electric field \((120 \ \mathrm{V/m}).\) Using these values, we can compute the intensity as follows: \(I = \frac{1}{2}(8.85 \times 10^{-12} \ \mathrm{C^2/N \cdot m^2})(3 \times 10^8 \ \mathrm{m/s})(120 \ \mathrm{V/m})^2 = 1.905 \times 10^{4} \ \mathrm{W/m^2}.\)

Compute the energy per photon and the number of photons per unit energy

The energy per photon can be obtained using \(E = h \cdot f,\) where \(h\) is the Planck's constant \((6.63 \times 10^{-34} \ \mathrm{J \cdot s}),\) and \(f\) can be calculated using the formula \(f = \frac{c}{\lambda},\) where \(\lambda\) is the wavelength of the laser \((640 \times 10^{-9} \ \mathrm{m}).\) So, we have \(f = \frac{3 \times 10^8 \ \mathrm{m/s}}{640 \times 10^{-9} \ \mathrm{m}} = 4.69 \times 10^{14} \ \mathrm{Hz}.\) The energy per photon is: \(E = (6.63 \times 10^{-34} \ \mathrm{J \cdot s})(4.69 \times 10^{14} \ \mathrm{Hz}) = 3.11 \times 10^{-19} \ \mathrm{J}\)

Determine the energy within the beam

The energy of the beam per unit area is given by the intensity \(I = 1.905 \times 10^{4} \ \mathrm{W/m^2}.\) We need to find the total energy in the beam and for that, we calculate the cross-sectional area of the beam using \(A = \pi (\frac{D}{2})^2,\) where \(D\) is the diameter of the beam \((1.5 \times 10^{-3} \ \mathrm{m}).\) So, we have \(A = \pi (\frac{1.5 \times 10^{-3} \ \mathrm{m}}{2})^2 = 1.767 \times 10^{-6} \ \mathrm{m^2}.\) Now, the energy within the beam is obtained as: \(E_{beam} = I \cdot A = (1.905 \times 10^{4} \ \mathrm{W/m^2})(1.767 \times 10^{-6} \ \mathrm{m^2}) = 3.364 \times 10^{-2} \ \mathrm{W}.\)

Find the number of photons emitted per second

To calculate the number of photons emitted per second, we divide the energy in the beam by the energy per photon: \(N = \frac{E_{beam}}{E} = \frac{3.364 \times 10^{-2} \ \mathrm{W}}{3.11 \times 10^{-19} \ \mathrm{J}} = 1.082 \times 10^{17} \ \mathrm{photons/s}.\) So, the laser emits approximately \(1.082 \times 10^{17}\) photons per second.