Solve because the probability is very small we can

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Solve: Because the probability is very small, we can write the decay rate 1 0.010 0.10 ns 0.10 ns P r t = = = Δ The lifetime of the excited state is 1 1 1 10 ns 0.10 ns r τ = = =
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42.17. Solve: The interval t = 0.50 ns is very small in comparison with the lifetime τ = 25 ns, so we can write Prob(decay in t ) = r t where r is the decay rate. The decay rate is related to the lifetime by r = 1/ τ = 1/(25 ns) = 0.040 ns –1 . Thus Prob(decay in t ) = (0.040 ns –1 )(0.50 ns) = 0.020 = 2.0%
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42.18. Solve: (a) If there are initially N 0 atoms in a state with lifetime τ , then the number remaining at time t is N exc = N 0 e t / τ . From Table 42.3, the 3 p state has a lifetime τ = 17 ns. Hence, N exc = 1.0 × 10 6 e 10/17 = 555,000 (b) Likewise at t = 30 ns, N exc = 1.0 × 10 6 e 30/17 = 171,000. (c) For t = 100 ns, N exc = 1.0 × 10 6 e 100/17 = 3000.
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42.19. Solve: (a) The number of atoms that have undergone a quantum jump to the ground state is 0.90 N 0 = 0.90(1.0 × 10 6 ) = 9.0 × 10 5 Because each transition is accompanied by a photon, the number of emitted photons is also 9.0 × 10 5 . (b) 10% of the atoms remain excited at t = 20 ns. Thus ( ) / 20 ns/ exc 0 0 0 20 ns 0.10 ln 0.10 8.7 ns t N N e N N e τ τ τ τ = = = =
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42.20. Visualize: We’ll first use Equation 42.18 to find the time t when 80% of the excited sodium atoms have decayed (that is, when 20% of the sodium atoms are still excited). Then we’ll take that t and use it again in Equa-tion 42.18 to find how many potassium atoms are still excited; t is the same in both equations. We look up the two lifetimes in Table 42.3: Na 17 ns τ = and P 26 ns. τ = We are given 8 0 1 0 10 N = . × for both kinds of atoms. Solve: Solve Equation 42.18 for t and plug in the values for sodium. exc 0 ln (17 ns)ln(0 20) 27 4 ns N t N τ = − = − . = . Now use this t to find exc N for potassium. 8 27 4 ns 26 ns 7 exc 0 (1 0 10 ) 3 5 10 t N N e e τ − / . / = = . × = . × Assess: 35% of the potassium atoms are still excited when only 20% of the sodium ones are; this seems reasonable.
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42.21. Solve: Since 1 mW = 0.001 W, the laser emits E light = 0.001 J of light energy per second. This energy consists of N photons. The energy of each photon is 19 photon 3.14 10 J hc E hf λ = = = × Because E light = NE photon , the number of photons is light 15 19 photon 0.001 J 3.2 10 3.13 10 J E N E = = = × × So, photons are emitted at the rate 3.2 × 10 15 s 1 .
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42.22. Solve: The energy of each photon is 20 photon 1.876 10 J hc E hf λ = = = × Because E light = NE photon , 22 20 light (5.0 10 )(1.876 10 J) 940 J E = × × = So, the energy output is 940 J every second, or 940 W.
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42.23. Solve: (a) The wavelength is 1240 eV nm 1060 nm 1.06 m 1.17 eV 0 eV hc E λ μ = = = = Δ (b) The energy per photon is E ph = 1.17 eV = 1.87 × 10 19 J. The power output of the laser is the number of photons per second times the energy per photon: P = (1.0 × 10 19 s 1 )(1.87 × 10 19 J) = 1.9 J/s = 1.9 W
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42.24. Solve: (a) For l = 3, the magnitude of the angular momentum vector is ( ) ( ) 1 3 3 1 12 L l l = + = + = = = = The angular momentum vector has a z -component L z = L cos θ along the z -axis. L z = m = where m is an integer between l and l , that is, between 3 and 3. The seven possible orientations of the angular momentum vector for l = 3 are shown in figure below.
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