Physics Solution Manual for 1100 and 2101

# 0 4 42 reasoning the mass m of the particle is

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Unformatted text preview: 29.3, we see that hf = hc λ = KE max + W0 (1) The work function W0 is a property of the metal surface itself, so it remains the same for any wavelength of incident light. When the wavelength of the incident light is λ1 = 221 nm, the maximum kinetic energy of the ejected electrons is KEmax,1 = 3.28×10−19 J, and when the wavelength is λ2, the maximum kinetic energy is twice as great: KEmax,2 = 2KEmax,1. We will use this information, with Equation (1), to determine the unknown wavelength λ2. SOLUTION Solving Equation (1) for the work function W0 yields W0 = − KE max + hc (2) λ Because the work function W0 isn’t affected by changing the wavelength of the incident light from λ1 to λ2, we have that W0 = −KEmax, 2 + hc λ2 = −KE max,1 + hc λ1 KE max, 2 − KE max,1 + or hc λ1 = hc λ2 (3) Solving Equation (3) for λ2 and substituting KEmax,2 = 2 KEmax,1, we obtain λ2 = hc KEmax, 2 − KEmax,1 + hc λ1 = hc 2KEmax, 1 − KEmax, 1 + hc λ1 = 1 KEmax, 1 hc + 1 (4) λ1 In the last step of Equation (4), we have divided both the numerator and the denominator by the product hc. Substituting the given values the into Equation (4), we find that Chapter 29 Problems λ2 = 1 ( 3.28 ×10 J ) 1 + −34 8 ( 6.63 ×10 J ⋅ s)( 3.00 ×10 m/s) 221 ×10−9 m −19 1517 = 1.62 × 10−7 m = 162 nm 11. REASONING The wavelength λ of the photon is related to its frequency f by λ = c/f (Equation 16.1), where c is the speed of light. The frequency of the photon is proportional to its energy E via f = E/h (Equation 29.2), where h is Planck’s constant. Thus, λ = ch/E. The photon energy is equal to the sum of the maximum kinetic energy KEmax of the ejected electron and the work function W0 of the metal; E = KE max + W0 (Equation 29.3). Substituting this expression for E into λ = ch/E gives λ= ch KE max + W0 (1) 2 The maximum kinetic energy is related to the maximum speed vmax by KE max = 1 mvmax 2 (Equation 6.2), where m is the mass of the electron. 2 SOLUTION Substituting KE max = 1 mvmax into Equation (1), and converting the work 2 function from electron-volts to joules, gives λ= ch 1 mv 2 max 2 + W0 ( 3.00 × 108 = 1 2 ( 9.11× 10−31 m/s )( 6.63 × 10−34 J ⋅ s ) = 1.9 × 10−7 m −19 2 1.60 × 10 J kg ) (1.2 × 106 m/s ) + ( 2.3 eV ) 1 eV 12. REASONING The total energy Q delivered by N photons is NE, where E is the energy carried by one photon, so that N = Q/E. Equation 29.2 indicates that the photon energy is E = hf, where h is Planck’s constant and f is the frequency. Thus, the number N of photons can be written as QQ N= = (1) E hf Equation 16.1 relates the frequency f of a photon to its wavelength λ according to f = c/λ, where c is the speed of light in a vacuum. Therefore, Equation (1) can be expressed as 1518 PARTICLES AND WAVES N= Q λQ = hf hc (2) According to Equation 12.4, the heat Q required to raise the temperature of a substance by an amount ∆T is Q = cspecific heatm∆T, where cspecific heat is the specific heat capacity of the substance and m is its mass. SOLUTION Combining Equation 12.4 with Equation (2), the number of photons required to raise the temperature by an amount ∆T is N= λQ or hc N= λ cspecific heat m ∆T hc Applying this result to each type of photon, we obtain N infrared = N blue = ( 6.0 × 10−5 m ) 840 J/ ( kg ⋅ C° ) ( 0.50 kg ) ( 2.0 C°) = ( 6.63 × 10−34 J ⋅ s )( 3.00 × 108 m/s ) ( 4.7 × 10−7 m ) 840 J/ ( kg ⋅ C° ) ( 0.50 kg ) ( 2.0 C°) = ( 6.63 × 10−34 J ⋅ s )( 3.00 × 108 m/s ) 2.5 × 1023 2.0 × 1021 13. SSM WWW REASONING AND SOLUTION a. According to Equation 24.5b, the electric field can be found from E = S / ( ε 0 c ) . The intensity S of the beam is Energy per unit time Nh f Nh c S= = = A A A λ = (1.30 × 1018 photons/s )( 6.63 × 10−34 J ⋅ s ) 3.00 × 108 m/s 2 514.5 × 10−9 m π (1.00 × 10−3 m ) = 1.60 × 105 W/m 2 where N is the number of photons per second emitted. Then, E = S / ( ε 0c ) = 7760 N / C Chapter 29 Problems 1519 b. According to Equation 24.3, the average magnetic field is B = E/c = 2.59 ×10–5 T 14. REASONING The heat required to melt the ice is given by Q = mL f , where m is the mass of the ice and L f is the latent heat of fusion for water (see Section 12.8). Since, according to Equation 29.2, each photon carries an energy of E = hf , the energy content of N photons is E Total = Nhf . According to Equation 16.1, f = c / λ , so we have E Total = Nhc λ If we assume that all of the photon energy is used to melt the ice, then, E Total = Q , so that Nhc :λ E = mL f 4 Q Total This expression may be solved for N to determine the required number of photons. SOLUTION a. We find that N= mL f λ hc = (2.0 kg)(33.5 × 10 4 J / kg)(620 × 10 –9 m) = 2.1 × 10 24 photons –34 8 (6.63 × 10 J ⋅ s)(3.00 × 10 m / s) b. The number N ′ of molecules in 2.0-kg of water is N ′ = (2.0 kg) F 1 mol I F × 10 molecules I = 6.7 × 10 6.022 J G × 10 kg JG 1 mol K 18 H KH 23 –3 25 molecules Therefore, on average, the number of water molecules that one photon converts from t...
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