Physics 43 HW 10 Ch 39

E: 9, 11, 15, 16, 19 P: 40, 42, 44, 45

39.9. Solve: (a) Using Equation 39.4, the wavelength of the photon is

−15 8 cchc()4.14×× 10 eV s( 3 10 m/s) λ == = = =4.14 × 10−6 m = 4140 nm fEhE 0.30 eV This is infrared light. (b) Likewise, for an energy of 3.0 eV, the wavelength is λ = 414 nm and is in the visible region. (c) For an energy of 30.0 eV, the wavelength is λ = 41.4 nm and is in the ultraviolet region. Assess: Since E ∝ λ−1 , the higher the energy of the photon, the smaller its wavelength.

39.11. Solve: (a) From Equation 39.4, the energy of each photon is −34 6 −28 Ephoton = hf = (6.63 × 10 J s)(101 × 10 Hz) = 6.696 × 10 J The number of photons in 104 J is

4 Etotal 10 J 29 N ==−28 =×1.5 10 Ephoton 6.696× 10 J The antenna emits 1.5 × 1029 photons per second. (b) The number of photons emitted per second is so enormous that we couldn’t possibly recognize the effects of single photons. It’s safe to treat the broadcast as an electromagnetic wave.

39.15. Solve: The de Broglie wavelength is λ = h/mv. Thus, h 6.63× 10−34 J s v == =1456 m/s mλ ()9.11×× 10−−31 kg() 500 109 m

1 2 A potential difference of ΔV will raise the kinetic energy of a rest electron by 2 mv . Thus,

−31 2 1 mv2 (9.11× 10 kg)() 1456 m/s eVΔ= mv2 ⇒Δ= V = =6.0 × 10−6 V 22e 2() 1.6× 10−19 C Assess: A mere 6.0 × 10−6 V is able to increase an electron’s speed to 1456 m/s.

39.16. Solve: The de Broglie wavelength is λ = h/mv. Thus h 6.63× 10−34 J s v == =3.970 × 107 m/s mλ ()1.67×× 10−−27 kg() 10 1015 m So, the kinetic energy of the proton is

1122− 7 72 1 eV Kmv==()1.67 × 10 kg() 3.970 × 10 m/s ×−19 =8.2 MeV 22 1.610× J

39.19. Model: For a “particle in a box,” the energy is quantized. Solve: The energy of the n = 1 state is

2 −−34 9 2 hhchλ ()6.63×× 10 J s() 600 10 m EEL===⇒==()1 =0.427 nm 1 88mL2 photon λ mc 8() 9.11×× 10−31 kg() 3.0 108 m/s

39.40. Solve: (a) The stopping potential is hh Vf=−f stop ee0

A graph of Vstop versus frequency f should be linear with x-intercept f0 and slope he. Since the x-intercept is f0 = 1.0 × 1015 Hz, the work function is −15 15 E0 = hf0 = (4.14 × 10 eV s)(1.0 × 10 Hz) = 4.14 eV (b) The slope of the graph is ΔV 8.0 V − 0 V stop ==4.0× 10−15 V s Δ×−×f 3.0 1015 Hz 1.0 1015 Hz

Because the slope of the Vstop versus f graph is he, an experimental value of Planck’s constant is he=×()4.0 10−−−15 V s =×( 1.6 1019 C)( 4.0 × 1015 V s) =× 6.4 10−34 J s Assess: This value of the Planck’s constant is about 3.5% lower than the accepted value.

39.42. Visualize:

The frequency of the photons was obtained using the equation f = c/λ and is tabulated below. The figure is a graph of Vstop versus f. λ (nm) 500 450 400 350 300 250 f (Hz) 6 × 1014 6.67 × 1014 7.50 × 1014 8.57 × 1014 1.0 × 1015 1.20 × 1015

Vstop (V) 0.19 0.48 0.83 1.28 1.89 2.74

Solve: (a) According to Equation 39.8, a graph of Vstop versus frequency f should be linear. The x-intercept is the threshold frequency f0. The slope of the graph is h/e. We can see from the graph that the x-intercept is 14 f0 = 5.5 × 10 Hz ⇒ E0 = hf0 = 2.3 eV

The work function E0 identifies the metal as potassium. 14 14 (b) The graph rises ΔVstop = 3.0 V over a run of Δ f = (12.5 – 5.5) × 10 Hz = 7.0 × 10 Hz. The experimental slope is ΔV stop =×4.28 10−15 V/Hz Δ f Equating the slope to h/e, we find that the experimental value for h is h = (slope)e = 6.85 × 10−34 J s = 4.28 × 10−15 eV s Assess: This value of h is about 3% higher than the accepted value.

39.44. Model: Electrons have both particle-like and wave-like properties. Visualize:

Please refer to Figure 39.13. 1 2−15 Solve: (a) The kinetic energy is Kmv==2 50 keV =× 8.0 10 J. Using this formula, the electron’s speed is

−15 2K ()28.010( × J) v == =×≈×1.32 1088 m/s 1.3 10 m/s m 9.11× 10−31 kg

(b) From Equation 22.7, the fringe spacing in a double-slit interference experiment is Δy =λL/d, where d is the slit separation and L is the distance to the viewing screen. The wavelength of the electrons is their de Broglie wavelength. We have hh 6.63× 10−34 J s λ == = =5.5 × 10−12 m pmv()9.11×× 10−31 kg() 1.32 108 m/s

−12 λL (5.5× 10 m)( 1.0 m) ⇒Δy = = =5.5 × 10−6 m = 5.5 μ m d 1.0× 10−6 m

39.45. Model: Neutrons have both particle-like and wave-like properties. Visualize:

Solve: (a) The kinetic energy of the neutron is

1122−−7 2 23−4 Kmv==22()1.67 × 10 kg( 200 m/s) =× 3.34 10 J =× 2.1 10 eV (b) The de Broglie wavelength at this speed is h 6.63× 10−34 J s λ == =×2.0 10−9 m = 2.0 nm mv ()1.67× 10−27 kg() 200 m/s (c) From Equation 22.7, the fringe spacing in a double-slit interference experiment is Δy =λL/d, where d is the slit separation and L is the distance to the detector. From Figure P39.45, the spacing between the two peaks with m = ±1 (on either side of the central maximum) is 1.4 times as long as the length of the reference bar, which gives Δy = 70 μm. Thus, the distance from the slits to the detector was

−−45 dyΔ (1.0×× 10 m)( 7.0 10 m) L == =3.5 m λ 2.0× 10−9 m