1056_PartUniversity Physics Solution

1056_PartUniversity Physics Solution - Relativity (a) In...

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Relativity 37-19 37.77. (a) In the center of momentum frame, the two protons approach each other with equal velocities (since the protons have the same mass). After the collision, the two protons are at rest but now there are kaons as well. In this situation the kinetic energy of the protons must equal the total rest energy of the two kaons 2 cm p 2( γ 1) mc −= 2 k 2 k cm p γ 1 1.526. m m =+ = The velocity of a proton in the center of momentum frame is then 2 cm cm 2 cm γ 1 0.7554 . γ vc c == To get the velocity of this proton in the lab frame, we must use the Lorentz velocity transformations. This is the same as “hopping” into the proton that will be our target and asking what the velocity of the projectile proton is. Taking the lab frame to be the unprimed frame moving to the left, cm cm and uv v v (the velocity of the projectile proton in the center of momentum frame). 2 cm lab lab lab lab p 2 2 cm lab 2 2 2 21 0.9619 γ 3.658 ( γ 2494 MeV. 1 1 1 vu v vcK m c uv v v c c c ′ + === =− = + + (b) lab k 2494 MeV 2.526. 2 2(493.7 MeV) K m (c) The center of momentum case considered in part (a) is the same as this situation. Thus, the kinetic energy required is just twice the rest mass energy of the kaons. cm 2(493.7 MeV) 987.4 MeV. K This offers a substantial advantage over the fixed target experiment in part (b). It takes less energy to create two kaons in the proton center of momentum frame.
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38-1 P HOTONS , E LECTRONS , AND A TOMS 38.1. IDENTIFY and SET UP: The stopping potential V 0 is related to the frequency of the light by 0 h Vf ee φ =− . The slope of V 0 versus f is h/e . The value f th of f when 0 0 V = is related to by th hf = . EXECUTE: (a) From the graph, 15 th 1.25 10 Hz f . Therefore, with the value of h from part (b), th 4.8 eV hf == . (b) From the graph, the slope is 15 3.8 10 V s ×⋅ . 16 15 34 ( )(slope) (1.60 10 C)(3.8 10 V s) 6.1 10 J s he −− × × = × (c) No photoelectrons are produced for th ff < . (d) For a different metal f th and are different. The slope is h/e so would be the same, but the graph would be shifted right or left so it has a different intercept with the horizontal axis. EVALUATE: As the frequency f of the light is increased above f th the energy of the photons in the light increases and more energetic photons are produced. The work function we calculated is similar to that for gold or nickel. 38.2. IDENTIFY and SET UP: cf λ = relates frequency and wavelength and Eh f = relates energy and frequency for a photon. 8 3.00 10 m/s c . 16 1 eV 1.60 10 J . EXECUTE: (a) 8 14 9 5.94 10 Hz 505 10 m c f × = × × (b) 34 14 19 (6.626 10 J s)(5.94 10 Hz) 3.94 10 J 2.46 eV f × × = × = (c) 2 1 2 Km v = so 19 15 22 ( 3 . 9 4 1 0 J ) 9.1 mm/s 9.5 10 kg K v m × = × 38.3. 8 14 7 3.00 10 m s 5.77 10 Hz λ 5.20 10 m c f × = × × 34 27 7 27 8 19 6.63 10 J s 1.28 10 kg m s λ 5.20 10 m (1.28 10 kg m s) (3.00 10 m s) 3.84 10 J 2.40 eV.
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1056_PartUniversity Physics Solution - Relativity (a) In...

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