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Chapter 38: Problem 47

A ruby laser consists mostly of alumina \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)\) and a small amount of chromium ions, responsible for its red color. One such laser of power \(3.00 \mathrm{~kW}\) emits light pulse of duration \(10.0 \mathrm{~ns}\) and of wavelength \(685 \mathrm{nm}\). a) What is the energy of the photons in the pulse? b) Determine the number of chromium atoms undergoing stimulated emission to produce this pulse.

Short Answer

Expert verified
Answer: The energy of the photons in the pulse is \(2.90 \times 10^{-19}\mathrm{J}\) and the number of chromium atoms undergoing stimulated emission to produce this pulse is approximately \(1.03 \times 10^{14}\) atoms.

Step by step solution

01

Calculate the frequency of the light

To find the energy of the photons, we first need to determine the frequency of the light. We can use the relationship between the speed of light, wavelength, and frequency: \(f = \dfrac{c}{\lambda}\) where c = 3.00 × 10^8 m/s (speed of light) and λ = 685 nm (wavelength) Convert wavelength to meters: \(\lambda = 685 \times 10^{-9} \mathrm{m}\) Now, plug in the values and calculate the frequency: \(f = \dfrac{3.00 \times 10^{8}\mathrm{m/s}}{685 \times 10^{-9}\mathrm{m}}\) \(f = 4.38 \times 10^{14}\mathrm{Hz}\)
02

Calculate the energy of the photons

Now, we can use the formula \(E = hf\) to calculate the energy of the photons: \(E = h \times f\) where h = 6.626 × 10^-34 Js (Planck's constant), and f = 4.38 × 10^14 Hz Now, plug in the values and calculate the energy of the photons: \(E = (6.626 \times 10^{-34}\mathrm{Js})(4.38 \times 10^{14}\mathrm{Hz})\) \(E = 2.90 \times 10^{-19}\mathrm{J}\)
03

Calculate the energy of the pulse

Using the given power and duration, we can find the energy of the pulse using the formula \(E_{pulse} = P \times t\): \(E_{pulse} = P \times t\) where P = 3.00 kW (power) and t = 10.0 ns (duration) Convert power and duration to SI units: \(P = 3.00 \times 10^{3}\mathrm{W}\), \(t = 10.0 \times 10^{-9}\mathrm{s}\) Now, plug in the values and calculate the energy of the pulse: \(E_{pulse} = (3.00 \times 10^{3}\mathrm{W})(10.0 \times 10^{-9}\mathrm{s})\) \(E_{pulse} = 3.00 \times 10^{-5}\mathrm{J}\)
04

Calculate the number of chromium atoms undergoing stimulated emission

Finally, we can find the number of chromium atoms undergoing stimulated emission using the formula \(N = \dfrac{E_{pulse}}{E_{photon}}\): \(N = \dfrac{E_{pulse}}{E_{photon}}\) where \(E_{pulse} = 3.00 \times 10^{-5}\mathrm{J}\) and \(E_{photon} = 2.90 \times 10^{-19}\mathrm{J}\) Now, plug in the values and calculate the number of chromium atoms undergoing stimulated emission: \(N = \dfrac{3.00 \times 10^{-5}\mathrm{J}}{2.90 \times 10^{-19}\mathrm{J}}\) \(N \approx 1.03 \times 10^{14}\) So, the energy of the photons in the pulse is \(2.90 \times 10^{-19}\mathrm{J}\) and the number of chromium atoms undergoing stimulated emission to produce this pulse is approximately \(1.03 \times 10^{14}\) atoms.

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