c. Assuming this value for , what is the energy of the proton, in electron... · OlympiadBench #1443

Physics · modern-physics

Problem

c. Assuming this value for

E_{b}

, what is the energy of the proton, in electron volts, that will allow the above reaction? This sets an upper limit on the energy of cosmic rays. The mass of the proton is given by

m_{p} c^{2}=938 \mathrm{MeV}

and the mass of the

\Delta

is given by

m_{\Delta} c^{2}=1232 \mathrm{MeV}

. Given constants:

g=9.8 \mathrm{~N} / \mathrm{kg}

k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}

c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}

N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}

\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)

1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}

m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}

m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}

\sin \theta \approx \theta-\frac{1}{6} \theta^{3}

for

|\theta| \ll 1

\begin{aligned} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ & e=1.602 \times 10^{-19} \mathrm{C} \\ & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ & \ln (1+x) \approx x \text { for }|x| \ll 1 \\ & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{aligned}

\begin{array}{lrl} \text { velocity parameter } & \beta & =\frac{v}{c} \\ \text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\ \text { relativistic momentum } & p & =\gamma \beta m c \\ \text { relativistic energy } & E & =\gamma m c^{2} \\ \text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}} \end{array}

Extra Supplementary Reading Materials: Relativistic particles obey the mass energy relation

E^{2}=(p c)^{2}+\left(m c^{2}\right)^{2}

where

E

is the relativistic energy of the particle,

p

is the relativistic momentum,

m

is the mass, and

c

is the speed of light. A proton with mass

m_{p}

and energy

E_{p}

collides head on with a photon which is massless and has energy

E_{b}

. The two combine and form a new particle with mass

m_{\Delta}

called

\Delta

, or "delta". It is a one dimensional collision that conserves both relativistic energy and relativistic momentum. Context question: a. Determine

E_{p}

in terms of

m_{p}, m_{\Delta}

, and

E_{b}

. You may assume that

E_{b}

is small. Context answer:

\frac{m_{\Delta}^{2}-m_{p}^{2}}{4 E_{b}}

Context question: b. In this case, the photon energy

E_{b}

is that of the cosmic background radiation, which is an EM wave with wavelength

1.06 \mathrm{~mm}

. Determine the energy of the photons, writing your answer in electron volts. Context answer:

0.00112

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