a. Determine in terms of , and . · OlympiadBench #1441

Physics · modern-physics

Problem

a. Determine

E_{p}

in terms of

m_{p}, m_{\Delta}

, and

E_{b}

. You may assume that

E_{b}

is small. 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.

Intuitive explanation

This is a relativistic collision problem where you need to apply conservation of energy and momentum to a head-on proton-photon interaction. Since $E_b$ is small compared to the rest energies involved, you can use the binomial approximation $(1+x)^n \approx 1 + nx$ to simplify the algebra. Set up the two conservation equations (energy and momentum, being careful about the head-on geometry), then solve for $E_p$ by eliminating the delta particle's momentum using the mass-energy relation.

Intuitive explanation

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