Fundamental Particles Problems and Solutions2

 Problems#1

A proton and an antiproton annihilate, producing two photons. Find the energy, frequency, and wavelength of each photon (a) if the p and are initially at rest and (b) if the p and collide head-on, each with an initial kinetic energy of 830 MeV.

Answer:

The rest energy of a proton or antiproton is 938.3 MeV. Conservation of linear momentum
requires that the two photons have equal energies

(a) The energy will be the proton rest energy, 938.3 MeV, corresponding to a frequency of
2.27 x 10²³ Hz × and a wavelength of 1.32 x 10^-15 m.

(b) The energy of each photon will be 938.3 MeV + 830 MeV = 1768 MeV with frequency 42.8 x 10²² Hz and wavelength 7.02 x 10 m.

Problem#2
For the nuclear reaction given in ₀n¹ + ¹⁰B₅ → ⁷Li₃ + ⁴He₂ assume that the initial kinetic energy and momentum of the reacting particles are negligible. Calculate the speed of the particle immediately after it leaves the reaction region.

Answer:
The mass decrease in the reaction is

∆m = m_n + m_B – m_Li – m_He

∆m = 1.008665u + 10.012937u – 7.016004u – 4.002603u = 0.002995u

and the energy released is

E = ∆mc² = 0.002995uc²

E = 0.002995(931.5 MeV/u) = 2.79 MeV

Assuming the initial momentum is zero,

m_Liv_Li = m_Hev_He

v_Li = (4.002603uv_He)/(7.016004u) = 0.5704961v_He              (1)

and
½ m_Liv_Li² + ½ m_Hev_He² = E

Where:
m_He = 4.002603u – 2(0.0005486 u) = 4.0015u = 6.645 x 10^-27kg

m_Li = 7.016004u – 3(0.0005486 u) = 7.0144u = 1.165 x 10^-26 kg

then, from (1)

(1.165 x 10^-26 kg)(0.5704961v_He)² + (6.645 x 10^-27kg)v_He² = 2(2.79 MeV)

(1.0436 x 10^-26 kg)v_He² = 8.939 x 10^-13J

v_He = 9.26 x 10⁶ m/s

Problem#3
Estimate the range of the force mediated by an meson that has mass 783 MeV/c².

Answer:
The range is limited by the lifetime of the particle, which itself is limited by the uncertainty principle.
∆E.∆t = h/4π

783 x 10⁶ x 1.602 x 10^-19 J.∆t = (6.62 x 10^-34 Js)/4π

∆t = 4.20 x 10^-25 s

The range of the force is

c∆t = (2.998 x 10⁸ m/s)(4.20 x 10^-25 s) = 1.26 x 10^-16 m = 0.126 fm

Problem#4
The starship Enterprise, of television and movie fame, is powered by combining matter and antimatter. If the entire 400-kg antimatter fuel supply of the Enterprise combines with matter,
how much energy is released? How does this compare to the U.S. yearly energy use, which is roughly 1.0 x 10²⁰J?

Answer:
The energy of the matter is

E = ∆mc² 
E = (400kg + 400 kg)(2.998 x 10⁸ m/s)² = 7.2 x 10¹⁹ J

Energy use in the U.S = (7.2 x 10¹⁹J) x 100%/(1.0 x 10²⁰ J) = 72%  

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