Instructor solution manual to accompany physical chemistry 7th ed by peter atkins chap11

Part 2: Structure 11 Quantum theory: introduction and principles Solutions to exercises Discussion questions E11.1(b) A successful theory of black-body radiation must be able to explain the energy density distribution of the radiation as a function of wavelength, in particular, the observed drop to zero as λ → Classical theory predicts the opposite However, if we assume, as did Planck, that the energy of the oscillators that constitute electromagnetic radiation are quantized according to the relation E = nhν = nhc/λ, we see that at short wavelengths the energy of the oscillators is very large This energy is too large for the walls to supply it, so the short-wavelength oscillators remain unexcited The effect of quantization is to reduce the contribution to the total energy emitted by the black-body from the high-energy short-wavelength oscillators, for they cannot be sufficiently excited with the energy available E11.2(b) In quantum mechanics all dynamical properties of a physical system have associated with them a corresponding operator The system itself is described by a wavefunction The observable properties of the system can be obtained in one of two ways from the wavefunction depending upon whether or not the wavefunction is an eigenfunction of the operator When the function representing the state of the system is an eigenfunction of the operator , we solve the eigenvalue equation (eqn 11.30) =ω in order to obtain the observable values, ω, of the dynamical properties When the function is not an eigenfunction of , we can only find the average or expectation value of dynamical properties by performing the integration shown in eqn 11.39 = E11.3(b) ∗ dτ No answer Numerical exercises E11.4(b) The power is equal to the excitance M times the emitting area P = MA = σ T (2πrl) = (5.67 × 10−8 W m−2 K −4 ) × (3300 K)4 × (2π ) × (0.12 × 10−3 m) × (5.0 × 10−2 m) = 2.5 × 102 W Comment This could be a 250 W incandescent light bulb E11.5(b) Wien’s displacement law is T λmax = c2 /5 E11.6(b) so λmax = 1.44 × 10−2 m K c2 = = 1.15 × 10−6 m = 1.15 µm 5T 5(2500 K) The de Broglie relation is λ = h h = p mv so v= v = 1.3 × 10−5 m s−1 h 6.626 × 10−34 J s = mλ (1.675 × 10−27 kg) × (3.0 × 10−2 m) INSTRUCTOR’S MANUAL 172 E11.7(b) The de Broglie relation is λ = h h = p mv so v= h 6.626 × 10−34 J s = mλ (9.11 × 10−31 kg) × (0.45 × 10−9 m) v = 1.6 × 106 m s−1 E11.8(b) The momentum of a photon is p= 6.626 × 10−34 J s h = = 1.89 × 10−27 kg m s−1 λ 350 × 10−9 m The momentum of a particle is p = mv so v= 1.89 × 10−27 kg m s−1 p = m 2(1.0078 × 10−3 kg mol−1 /6.022 × 1023 mol−1 ) v = 0.565 m s−1 E11.9(b) The energy of the photon is equal to the ionization energy plus the kinetic energy of the ejected electron Ephoton = Eionize + Eelectron and λ = so hc = Eionize + 21 mv λ (6.626 × 10−34 J s) × (2.998 × 108 m s−1 ) = Eionize + 21 mv 5.12 × 10−18 J + 21 (9.11 × 10−31 kg) × (345 × 103 m s−1 )2 hc = 3.48 × 10−8 m = 38.4 nm E11.10(b) The uncertainty principle is ¯ p x ≥ 21 h so the minimum uncertainty in position is x = h ¯ p = h ¯ 1.0546 × 10−34 J s = 2m v 2(9.11 × 10−31 kg) × (0.000 010) × (995 × 103 m s−1 ) = 5.8 × 10−6 m E11.11(b) E = hν = hc ; λ E(per mole) = NA E = NA hc λ hc = (6.62608 × 10−34 J s) × (2.99792 × 108 m s−1 ) = 1.986 × 10−25 J m NA hc = (6.02214 × 1023 mol−1 ) × (1.986 × 10−25 J m) = 0.1196 J m mol−1 1.986 × 10−25 J m 0.1196 J m mol−1 ; E(per mole) = λ λ We can therefore draw up the following table Thus, E = QUANTUM THEORY: INTRODUCTION AND PRINCIPLES E/J E/(kJ mol−1 ) 9.93 × 10−19 1.32 × 10−15 1.99 × 10−23 598 7.98 × 105 0.012 λ (a) 200 nm (b) 150 pm (c) 1.00 cm 173 E11.12(b) Assuming that the He atom is free and stationary, if a photon is absorbed, the atom acquires its momentum p, achieving a speed v such that p = mv p v= m = 4.00 × 1.6605 × 10−27 kg = 6.642¯ × 10−27 kg m h p= λ 6.626 × 10−34 J s = 3.313¯ × 10−27 kg m s−1 200 × 10−9 m 3.313¯ × 10−27 kg m s−1 p = = 0.499 m s−1 v= m 6.642 × 10−27 kg p= (a) 6.626 × 10−34 J s = 4.417¯ × 10−24 kg m s−1 150 × 10−12 m 4.417¯ × 10−24 kg m s−1 p = 665 m s−1 = v= m 6.642 × 10−27 kg (b) p= (c) p= 6.626 × 10−34 J s = 6.626 × 10−32 kg m s−1 1.00 × 10−2 m p 6.626 × 10−32 kg m s−1 = 9.98 × 10−6 m s−1 v= = m 6.642 × 10−27 kg E11.13(b) Each emitted photon increases the momentum of the rocket by h/λ The final momentum of the Nh rocket will be N h/λ, where N is the number of photons emitted, so the final speed will be λmrocket The rate of photon emission is the power (rate of energy emission) divided by the energy per photon (hc/λ), so N = tP λ hc v (10.0 yr) × (365 day yr −1 ) × (24 h day−1 ) × (3600 s h−1 ) × (1.50 × 103 W) (2.998 × 108 m s−1 ) × (10.0 kg) = and v= tP λ hc × h λmrocket = tP cmrocket = 158 m s−1 E11.14(b) Rate of photon emission is rate of energy emission (power) divided by energy per photon (hc/λ) (a) (b) (0.10 W) × (700 × 10−9 m) Pλ = 3.52 × 1017 s−1 = hc (6.626 × 10−34 J s) × (2.998 × 108 m s−1 ) (1.0 W) × (700 × 10−9 m) = 3.52 × 1018 s−1 rate = (6.626 × 10−34 J s) × (2.998 × 108 m s−1 ) rate = E11.15(b) Wien’s displacement law is T λmax = c2 /5 so T = c2 1.44 × 10−2 m K = 1800 K = 5λmax 5(1600 × 10−9 m) INSTRUCTOR’S MANUAL 174 E11.16(b) Conservation of energy requires Ephoton = + EK = hν = hc/λ so and EK = 21 me v so v = (a) (b) EK = hc/λ − 2EK 1/2 me (6.626 × 10−34 J s) × (2.998 × 108 m s−1 ) − (2.09 eV) × (1.60 × 10−19 J eV−1 ) 650 × 10−9 m But this expression is negative, which is unphysical There is no kinetic energy or velocity because the photon does not have enough energy to dislodge the electron EK = EK = (6.626 × 10−34 J s) × (2.998 × 108 m s−1 ) − (2.09 eV) × (1.60 × 10−19 J eV−1 ) 195 × 10−9 m = 6.84 × 10−19 J and v = E11.17(b) E = hν = h/τ , 2(3.20 × 10−19 J) 9.11 × 10−31 kg 1/2 = 1.23 × 106 m s−1 so (a) E = 6.626 × 10−34 J s/2.50 × 10−15 s = 2.65 × 10−19 J = 160 kJ mol−1 (b) E = 6.626 × 10−34 J s/2.21 × 10−15 s = 3.00 × 10−19 J = 181 kJ mol−1 (c) E = 6.626 × 10−34 J s/1.0 × 10−3 s = 6.62 × 10−31 J = 4.0 × 10−10 kJ mol−1 E11.18(b) The de Broglie wavelength is λ= h p The momentum is related to the kinetic energy by EK = p2 2m so p = (2mEK )1/2 The kinetic energy of an electron accelerated through V is eV = 1.60 × 10−19 J, so λ= h (2mEK )1/2 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES (a) λ= 175 6.626 × 10−34 J s (2(9.11 × 10−31 kg) × (100 eV) × (1.60 × 10−19 J eV−1 ))1/2 = 1.23 × 10−10 m (b) λ= 6.626 × 10−34 J s (2(9.11 × 10−31 kg) × (1.0 × 103 eV) × (1.60 × 10−19 J eV−1 ))1/2 = 3.9 × 10−11 m (c) λ= 6.626 × 10−34 J s (2(9.11 × 10−31 kg) × (100 × 103 eV) × (1.60 × 10−19 J eV−1 ))1/2 = 3.88 × 10−12 m E11.19(b) The minimum uncertainty in position is 100 pm Therefore, since p≥ v= h ¯ x = ¯ x p ≥ 21 h 1.0546 × 10−34 J s = 5.3 × 10−25 kg m s−1 2(100 × 10−12 m) 5.3 × 10−25 kg m s−1 p = = 5.8 × 10−5 m s−1 m 9.11 × 10−31 kg E11.20(b) Conservation of energy requires Ephoton = Ebinding + 21 me v = hν = hc/λ and Ebinding = so Ebinding = hc/λ − 21 me v (6.626 × 10−34 J s) × (2.998 × 108 m s−1 ) 121 × 10−12 m − (9.11 × 10−31 kg) × (5.69 × 107 m s−1 )2 = 1.67 × 10−16 J Comment This calculation uses the non-relativistic kinetic energy, which is only about per cent less than the accurate (relativistic) value of 1.52 × 10−15 J In this exercise, however, Ebinding is a small difference of two larger numbers, so a small error in the kinetic energy results in a larger error in Ebinding : the accurate value is Ebinding = 1.26 × 10−16 J Solutions to problems Solutions to numerical problems P11.3 hν J s × s−1 , [θE ] = =K k J K−1 In terms of θE the Einstein equation [11.9] for the heat capacity of solids is θE = CV = 3R θE × T eθE /2T eθE /T − , classical value = 3R hν as demonstrated in the text kT θE (Section 11.1) The criterion for classical behaviour is therefore that T It reverts to the classical value when T θE = θE or when hν (6.626 × 10−34 J Hz−1 ) × ν = 4.798 × 10−11 (ν/Hz)K = k 1.381 × 10−23 J K−1 INSTRUCTOR’S MANUAL 176 (a) For ν = 4.65 × 1013 Hz, θE = (4.798 × 10−11 ) × (4.65 × 1013 K) = 2231 K (b) For ν = 7.15 × 1012 Hz, θE = (4.798 × 10−11 ) × (7.15 × 1012 K) = 343 K Hence (a) CV = 3R 2231 K × 298 K (b) CV = 3R 343 K × 298 K ¯ e2231/(2×298) e2231/298 − e343/(2×298) e343/298 − = 0.031 = 0.897 Comment For many metals the classical value is approached at room temperature; consequently, the failure of classical theory became apparent only after methods for achieving temperatures well below 25◦ C were developed in the latter part of the nineteenth century P11.5 The hydrogen atom wavefunctions are obtained from the solution of the Schrăodinger equation in Chapter 13 Here we need only the wavefunction which is provided It is the square of the wavefunction that is related to the probability (Section 11.4) ψ2 = πa03 e−2r/a0 , δτ = πr , r0 = 1.0 pm If we assume that the volume δτ is so small that ψ does not vary within it, the probability is given by ψ δτ = 4r03 −2r/a0 e = × 3 3a0 ψ δτ = (a) r=0: (b) r = a0 : 4 ψ δτ = 1.0 −2r/a0 e 53 1.0 = 9.0 × 10−6 53 1.0 −2 e = 1.2 × 10−6 53 Question If there is a nonzero probability that the electron can be found at r = how does it avoid destruction at the nucleus? (Hint See Chapter 13 for part of the solution to this difficult question.) P11.7 According to the uncertainty principle, p q ≥ 21 h ¯, where q and p are root-mean-square deviations: q = ( x − x )1/2 and p = ( p − p )1/2 To verify whether the relationship holds for the particle in a state whose wavefunction is = (2a/π )1/4 e−ax , We need the quantum-mechanical averages x , x , p , and p ∞ x = ∗ x dτ = ∞− 2a 1/4 −ax e x π 2a 1/4 −ax e dx, π QUANTUM THEORY: INTRODUCTION AND PRINCIPLES 177 ∞ 2a 1/2 xe−2ax dx = 0; π x = −∞ ∞ x 2a 1/4 −ax 2 e x π = −∞ 1/2 2a π x2 = q = so ∞− π 1/2 = ; 3/2 4a 2(2a) 2a 1/2 ∞ p ∞ 2a 1/2 x e−2ax dx, π 2a 1/4 −ax e dx = π = ∞ h ¯ d i dx ∗ −∞ dx and p ∗ = −∞ −¯h2 d2 dx dx We need to evaluate the derivatives: 2a 1/4 (−2ax)e−ax π d = dx and 2 2a 1/4 [(−2ax)2 e−ax + (−2a)e−ax ] = π d2 = dx ∞ So p = −∞ 2a 1/4 −ax e π 2¯h =− i ∞ p = −∞ p h ¯ i 2a 1/4 (4a x − 2a)e−ax π 2a 1/4 (−2ax)e−ax dx π ∞ 2a 1/2 xe−2ax dx = 0; π −∞ 2a 1/4 −ax 2a 1/4 e (−¯h2 ) (4a x − 2a)e−ax dx, π π ∞ 2a 1/2 = (−2a¯h ) (2ax − 1)e−2ax dx, π −∞ p2 and = (−2a¯h2 ) 2a π 1/2 2a π 1/2 π 1/2 − 2(2a)3/2 (2a)1/2 = a¯h2 ; p = a 1/2 h ¯ × a 1/2 h ¯ = 1/2¯h, 2a 1/2 which is the minimum product consistent with the uncertainty principle Finally, q p= INSTRUCTOR’S MANUAL 178 Solutions to theoretical problems P11.9 We look for the value of λ at which ρ is a maximum, using (as appropriate) the short-wavelength (high-frequency) approximation ρ= 8πhc λ5 ehc/λkT −1 ehc/λkT ρ=0 ehc/λkT − dρ hc =− ρ+ dλ λ λ kT Then, −5 + [11.5] at λ = λmax hc ehc/λkT =0 × hc/λkT λkT e −1 Hence, − 5ehc/λkT + hc hc/λkT e =0 λkT hc [short wavelengths, high frequencies], this expression simplifies We neglect the initial 5, λkT cancel the two exponents, and obtain If hc = 5λkT or λmax T = for λ = λmax and hc λkT hc = 2.88 mm K , in accord with observation 5k Comment Most experimental studies of black-body radiation have been done over a wavelength range of a factor of 10 to 100 of the wavelength of visible light and over a temperature range of 300 K to 10 000 K P11.10 Question Does the short-wavelength approximation apply over all of these ranges? Would it apply to the cosmic background radiation of the universe at 2.7 K where λmax ≈ 0.2 cm? 8πhc ρ= [11.5] hc/λkT e −1 λ hc decreases, and at very long wavelength hc/λkT λkT the exponential in a power series Let x = hc/λkT , then As λ increases, ex = + x + ρ= 8πhc λ5 x + x3 + · · · 2! 3! 1+x+ 2! x + 3! x + ··· − 1 8π hc 8πhc = 1+x−1 λ λ5 8πkT = λ4 lim ρ = λ→∞ This is the Rayleigh–Jeans law [11.3] hc/λkT Hence we can expand QUANTUM THEORY: INTRODUCTION AND PRINCIPLES P11.12 ρ= 179 8πhc [11.5] × hc/λkT e −1 λ ∂ρ 40πhc × =− ∂λ λ6 = 8πhc × λ5 λ = − ∂ρ =0 ∂λ ehc/λkT − 1 ehc/λkT − ×ρ 1− when λ = λmax =1 5λmax kT hc − e−hc/λmax kT =1 then ehc/λkT − ehc/λkT hc + λ λ kT ehc/λkT − and 1 − e−hc/λmax kT hc ; λmax kT × − × ehc/λkT × − λ2hckT hc −hc/λkT 5λkT − e hc 5λmax kT Let x = − 8π hc λ5 − e−x = x or = x − e−x The solution of this equation is x = 4.965 Then h = 4.965λmax kT c However M = σT = 2π k 15c2 h3 (1) T4 (2) Substituting (1) into (2) yields M≈ ≈ k≈ ≈ 2π k 15c2 × c T4 4.965λmax kT 2π ckT 1835.9λ3max 1835.9λ3max M 2π cT 1835.9(1.451 × 10−6 m)3 × (904.48 × 103 W) 2π (2.998 × 108 m s−1 ) × (2000 K) × (1.000 m2 ) k ≈ 1.382 × 10−23 J K−1 (3) INSTRUCTOR’S MANUAL 180 Substituting (3) into (1) h≈ 5(1.451 × 10−6 m) × (1.382 × 10−23 J K−1 ) × (2000 K) 2.998 × 108 m s−1 h ≈ 6.69 × 10−34 J s Comment These calculated values are very close to the currently accepted values for these constants P11.14 In each case form N ψ; integrate (N ψ)∗ (N ψ) dτ set the integral equal to and solve for N (a) ψ =N 2− r a0 ψ2 = N2 − ψ dτ = N e−r/2a0 r −r/a0 e a0 ∞ 4r − 4r r4 + a0 a0 = N × 2a03 − × hence N = π e−r/a0 dr 6a04 24a + 20 a0 a0 sin θ dθ 2π dφ × (2) × (2π ) = 32π a03 N ; 1/2 32πa03 where we have used ∞ (b) n! x n e−ax dx = n+1 [Problem 11.13] a ψ = N r sin θ cos φ e−r/(2a0 ) ψ dτ = N ∞ r e−r/a0 dr = N 4!a05 −1 π 0 sin2 θ sin θ dθ 2π cos2 φ dφ (1 − cos2 θ) d cos θ × π = N 4!a05 − where we have used π π = 32π a05 N02 ; cosn θ sin θ dθ = − −1 hence N= cosn θ d cos θ = 1/2 32π a05 −1 x n dx and the relations at the end of the solution to Problem 11.8 [See Student’s solutions manual.] QUANTUM THEORY: INTRODUCTION AND PRINCIPLES P11.16 P11.19 181 Operate on each function with i; if the function is regenerated multiplied by a constant, it is an eigenfunction of i and the constant is the eigenvalue (a) f = x − kx i(x − kx) = −x + kx = −f Therefore, f is an eigenfunction with eigenvalue, −1 (b) f = cos kx i cos kx = cos(−kx) = cos kx = f Therefore, f is an eigenfunction with eigenvalue, +1 (c) f = x + 3x − i(x + 3x − 1) = x − 3x − = constant × f Therefore, f is not an eigenfunction of i The kinetic energy operator, Tˆ , is obtained from the operator analogue of the classical equation EK = p2 2m that is, (p) ˆ Tˆ = 2m h ¯ d [11.32]; i dx pˆ x = pˆ x2 = −¯h2 hence d2 dx and h ¯ d2 Tˆ = − 2m dx Then T = N2 = ψ∗ −¯h2 2m ψ ∗ (−k ) × (eikx cos χ + e−ikx sin χ ) dτ ψ ∗ψ dτ h ¯ d [11.32] i dx px = N ψ ∗ pˆ x ψ dx; h ¯ = (a) dτ N2 = ψ ∗ψ dτ ψ ∗ ψ dτ −¯h2 px = ψ ∗ψ d2 ikx cos χ + e−ikx sin χ ) dτ ψ ∗ dx (e = 2m P11.20 pˆ ψ dτ ψ ∗ 2m pˆ x2 ψ dτ = 2m ψ ∗ pˆ x ψ dx i = ∗ ψ ψ dx ψ = eikx , N2 = ψ∗ ψ ∗ψ dψ dx ψ ∗ψ dx dψ = ikψ dx Hence, h ¯ × ik ψ ∗ ψ dx px = i = k¯h ψ ∗ ψ dx dx dτ = h ¯ k ψ ∗ ψ dτ h ¯ k2 = ∗ 2m 2m ψ ψ dτ INSTRUCTOR’S MANUAL 182 (b) ψ = cos kx, ∞ −∞ ψ∗ dψ = −k sin kx dx ∞ dψ dx = −k cos kx sin kx dx = dx −∞ Therefore, px = 2 dψ = −2αxe−αx (c) ψ = e−αx , dx ∞ −∞ ψ∗ ∞ dψ xe−2αx dx = dx = −2α dx −∞ [by symmetry, since x is an odd function] Therefore, px = P11.23 No solution Solution to applications P11.27 (a) Consider any infinitesimal volume element dx dy dz within the hemisphere (Figure 11.1) that has a radius equal to the distance traveled by light in the time dt (c dt) The objective is to find the total radiation flux perpendicular to the hemisphere face at its center Imagine an infinitesimal area A at that point Let r be the distance from dx dy dz to A and imagine the infinitesimal area A perpendicular to r E is the total isotropic energy density in dx dy dz E dx dy dz is the energy emitted in dt A /4π r is the fraction of this radiation that passes through A The radiation flux that originates from dx dy dz and passes through A in dt is given by: JA = A ✁ 4π r E dx dy dz = E dx dy dz 4π r dt A dt   The contribution of JA to the radiation flux through A, JA , is given by the expression JA × (  A cos θ )/  A = JA cos θ The integration of this expression over the whole hemisphere gives an JAЈ A c dt JA AЈ c dt dx dy dz Figure 11.1 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES 183 expression for JA Spherical coordinates facilitate to integration: dx dy dz = r sin θ dθ dφ dr = −r d(cos θ ) dφ dr where ≤ θ ≤ 2π and ≤ θ ≤ π/2 JA = E dx dy dz 4π r dt cos(θ ) hemisphere E cos(θ ) = ✓2 d(cos θ) dφ dr} {−r✓ dt r  4π  hemisphere cos(π/2) E =− 4π dt cos(θ ) d(cos θ) E ✟✚ 4π dt ✟ cE JA = − J = dφ cos(0) =− c dt 2π    w dw  (2✚ π ) (c ✚ dt) dr −1 (2✁) {Subscript “A” has been a bookkeeping device It may be dropped.} 2✁ cE dJ = or 8πhc dλ dE = hc/λRT λ (e − 1) c dE [eqn 11.5] By eqn 16.1 ν˜ = 1/λ Taking differentials to be positive, dν˜ = dλ/λ2 or dλ = λ2 dν˜ = dν˜ /˜ν The substitution of ν˜ for λ gives: 8πhcν˜ dν˜ dE = hcν˜ /kT e −1 2π hc2 ν˜ Thus, dJ = f(˜ν ) dν˜ where f(˜ν ) = hcν˜ /kT e −1 The value of the Stefan–Boltzmann constant σ is defined by the low n = 0∞ dJ (˜ν ) = σ T n is called the total exitance Let x = hcν˜ /kT (or ν˜ = kT x/ hc), substitute the above equation for dJ (˜ν ) into the Stefan–Boltzmann low, and integrate ∞ n = o = 2π k T 2πhc2 ν˜ dν˜ = h3 c ehcν˜ /kT − 2πk T h3 c π4 15 = ∞ x dx ex − 2π k 15 h3 c2 T4 2π k = 5.6704 × 10−8 W m−2 K −4 15h3 c2 The function f (˜ν ) gives radiation density in units that are compatible with those often used in discussions of infrared radiation which lies between about 33 cm−1 and 12 800 cm−1 (Fig 11.2) Thus, σ = INSTRUCTOR’S MANUAL 184 Blackbody radiation density at 288.16 K F( ) /10–11 J m–2 4.5 1.5 0 500 1000 1500 2000 2500 Wavenumber, /cm–1 3000 Figure 11.2 By graphing f (˜ν ) at the observed average temperature of the Earth’s surface (288.16 K) we easily see that the Earth’s black-body emissions are in the infrared with a maximum at about 600 cm−1 (b) Let R represent the radius of the Earth Assuming an average balance between the Earth’s absorption of solar radiation and Earth’s emission of black-body radiation into space gives: Solar energy absorbed = black-body energy lost πR (1 − albedo)(solar energy flux) = (4π R )(σ T ) Solving for T gives: T = = (1 − albedo)(solar energy flux) 1/4 4σ (1 − 0.29)(0.1353 W cm−2 ) 4(5.67 × 10−12 W cm−2 K −4 1/4 = 255 K This is an estimate of what the Earth’s temperature would be in the absence of the greenhouse effect  ... short-wavelength oscillators remain unexcited The effect of quantization is to reduce the contribution to the total energy emitted by the black-body from the high-energy short-wavelength oscillators, for they... is the minimum product consistent with the uncertainty principle Finally, q p= INSTRUCTOR S MANUAL 178 Solutions to theoretical problems P11.9 We look for the value of λ at which ρ is a maximum,... (Figure 11.1) that has a radius equal to the distance traveled by light in the time dt (c dt) The objective is to find the total radiation flux perpendicular to the hemisphere face at its center