Physics Solution Manual for 1100 and 2101

# 63 1034 j s h h 12 1036 m mv m 2ah 41 kg 2 98

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Unformatted text preview: hoton possesses a greater energy E, because E = hf (Equation 29.2), where h is Planck’s constant and f is the frequency. 4. (e) As the wavelength of a photon becomes smaller, its frequency and its energy become larger (see Section 29.3). Since the energy of an incident photon is equal to the maximum kinetic energy of the photoelectron plus the work function of the metal, increasing the incident energy increases the maximum kinetic energy of the photoelectron. The work function is a characteristic of the metal, and does not depend on the photon. 5. (d) The maximum kinetic energy of the ejected photoelectrons depends on the energy of the incident photons. Doubling the number of photons per second that strikes the surface does not change the energy of the incident photons, which depends only on the frequency of the photons. Thus, the maximum kinetic energy of the ejected photoelectrons does not change. However, doubling the number of photons per second that strikes the surface means that twice as many photoelectrons per second are ejected from the surface. 6. (b) Whether or not electrons are ejected from the surface of the metal depends on the energy of the incident photons and the work function of the metal The work function depends on the type of metal (e.g., aluminum or copper) from which the plate is made. (See Section 29.3). 7. W0 = 3.6 × 10−19 J 8. KEmax = 2.7 × 10−20 J 9. (d) According to the discussion in Section 29.4, a photon has a momentum whose magnitude p is related to its wavelength λ by p = h/λ. Chapter 29 Answers to Focus on Concepts Questions 1511 10. (b) In the Compton effect, an X-ray photon strikes an electron and, like two billiard balls (particles) colliding on a pool table, the X-ray photon scatters in one direction and the electron recoils in another direction after the collision. 11. (a) In the Compton effect, some of the energy of the incident photon is given to the recoil electron. Therefore, the energy of the scattered photon is less than that of the incident photon. Since the energy of a photon depends inversely on its wavelength (see Section 29.3), the wavelength of the scattered photon is greater than that of the incident photon. 12. λ ′ = 0.35 nm 13. (c) The de Broglie wavelength λ depends inversely on the magnitude p of the momentum; λ = h/p (Equation 29.8), where h is Planck’s constant. Particle A, having the smaller charge, has the smaller electric potential energy (see Section 19.2). Consequently, after accelerating through the potential difference, particle A has the smaller kinetic energy, and hence, the smaller momentum. Thus, particle A has the longer de Broglie wavelength. 14. (e) The de Broglie wavelength λ depends inversely on the magnitude p of the momentum; λ = h/p (Equation 29.8), where h is Planck’s constant. Therefore, as the momentum decreases, the wavelength increases, and vice versa. In A, the proton is moving opposite to the direction of the electric field, so the proton is slowing down, and its momentum is decreasing. In B, the proton is accelerating, and its momentum is increasing. In C, the proton moves parallel to the magnetic field. According to the discussion in Section 21.2, the proton does not experience a force, so its momentum remains constant. In D the proton is moving perpendicular to the direction of the magnetic field. According to the discussion in Section 21.3, such a situation does not change the magnitude of the proton’s momentum. 15. λ = 2.1 × 10−14 m 16. (b) According to the Heisenberg uncertainty principle, the uncertainty ∆y in a particle’s position is related to the uncertainty ∆py in its momentum by (see Equation 29.10) ∆y ≥ h / ( 4π ∆p y ) . If ∆py = 0 kg⋅m/s, then ∆y becomes infinitely large. 1512 PARTICLES AND WAVES CHAPTER 29 PARTICLES AND WAVES PROBLEMS 1. REASONING AND SOLUTION The energy of a single photon is E = hf = (6.63 × 10–34 J⋅s)(98.1 × 106 Hz) = 6.50 × 10–26 J The number of photons emitted per second is Power radiated 5.0 × 10 4 W = = Energy per photon 6.50 × 10 −26 J 2. 7 .7 × 10 29 photons / s REASONING The energy of a photon of frequency f is, according to Equation 29.2, E = hf , where h is Planck's constant. Since the frequency and wavelength are related by f = c / λ (see Equation 16.1), the energy of a photon can be written in terms of the wavelength as E = hc / λ . These expressions can be solved for both the wavelength and the frequency. SOLUTION a. The wavelength of the photon is λ= hc (6.63 × 10 –34 J ⋅ s)(3.00 × 10 8 m / s) = = 1.63 × 10 –7 m –18 E 1.22 × 10 J b. Using the answer from part (a), we find that the frequency of the photon is f= c λ = 3.00 × 10 8 m / s = 1.84 × 10 15 Hz 1.63 × 10 –7 m Alternatively, we could use Equation 29.2 directly to obtain the frequency: f= E 1.22 × 10 –18 J = = 1.84 × 10 15 Hz h 6.63 × 10 –34 J ⋅ s c. The wavelength and frequency values shown in Figure 24.9 indicate that this photon corresponds to electromagnetic radi...
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