The Laser Problems and Solutions 1

 Problem#1

PRK Surgery. Photorefractive keratectomy (PRK) is a laser-based surgical procedure that corrects near- and farsightedness by removing part of the lens of the eye to change its curvature and hence focal length. This procedure can remove layers 0.25 µm thick using pulses lasting 12.0 ns from a laser beam of wavelength 193 nm. Low-intensity beams can be used because each individual photon has enough energy to break the covalent bonds of the tissue. (a) In what part of the electromagnetic spectrum does this light lie? (b) What is the energy of a single photon? (c) If a 1.50-mW beam is used, how many photons are delivered to the lens in each pulse?

Answer:

Visible light has wavelengths from about 400 nm to about 700 nm. The energy of each photon is

E = hf = hc/λ = (1.99 x 10^-25 J.m)/λ

The power is the total energy per second and the total energy E_tot is the number of photons N times the energy E of each photon.

(a) 193 nm is shorter than visible light so is in the ultraviolet.

(b) E = hf = hc/λ = (1.99 x 10^-25 J.m)/(193 x 10^-9 m) = 1.03 x 10^-18 J = 6.44 eV

(c) P = E_tot/t = NE/t, so

N = Pt/E = (1.50 x 10^-3 W)(12.0 x 10^-9 s)/(1.03 x 10^-18 J) = 1.75 x 10⁷ photons

A very small amount of energy is delivered to the lens in each pulse, but this still corresponds to a large number of photons.

Problem#2

A large number of neon atoms are in thermal equilibrium. What is the ratio of the number of atoms in a 5s state to the number in a 3p state at (a) 300 K; (b) 600 K; (c) 1200 K? The energies of these states, relative to the ground state, E_5s = 20.66 eV are and E_3p = 18.70 eV. (d) At any of these temperatures, the rate at which a neon gas will spontaneously emit 632.8-nm radiation is quite low. Explain why.

Answer:

We use n_5s/n_3p = e^{-(E5s – E3p)/kT}

E_5s = 20.66 eV; E_3p = 18.70 eV and k = 1.38 x 10^-23 J/K

E_5s – E_3p = 20.66 eV – 18.70 eV = 1.96 eV (1.602 x 10^-19 J/1eV) = 3.140 x 10^-19 J

(a) T = 300 K

n_5s/n_3p = e^{-(E5s – E3p)/kT} = e^-75.79 = 1.2 x 10^-33

(b) T = 600 K

n_5s/n_3p = e^{-(E5s – E3p)/kT} = e^-37.90 = 3.5 x 10^-17

(c) T = 1200 K

n_5s/n_3p = e^{-(E5s – E3p)/kT} = e^-18.95 = 5.9 x 10^-9

(d) At each of these temperatures the number of atoms in the 5s excited state, the initial state for
the transition that emits 632.8 nm radiation, is quite small. The ratio increases as the temperature increases.

Problem#3

Figure 1 shows the energy levels of the sodium atom. The two lowest excited levels are shown in columns labeled ²P_3/2 and ²P_1/2. Find the ratio of the number of atoms in a state to the number in a ²P_3/2 state for a sodium gas in thermal equilibrium at 500 K. In which state are more atoms found?

Answer:

The energy of each of these excited states above the ground state is hc/λ where λ is the wavelength of the photon emitted in the transition from the excited state to the ground state.

We use (n₂P_3/2)/(n₁P_1/2) = e^{-(E2P3/2 – E2P1/2)/kT}

∆E_3/2-g = hc/λ₁ = (6.626 x 10^-34 J.s)(2.998 x 10⁸ m/s)/(5.890 x 10^-7 m) = 3.373 x 10^-19 J

∆E_1/2-g = hc/λ₂ = (6.626 x 10^-34 J.s)(2.998 x 10⁸ m/s)/(5.896 x 10^-7 m) = 3.369 x 10^-19 J

So ∆E_3/2-g – ∆E_1/2-g = 3.373 x 10^-19 J – 3.369 x 10^-19 J = 4.00 x 10^-22 J

Then (n₂P_3/2)/(n₁P_1/2) = 0.944. So more atoms are in the 2P_1/2 state.