Respuesta :
Given Information:
Kinetic energy of proton = KE = 5 MeV
Radius = R = 0530 m
Speed of light = c = 3x10⁸ m/s
Permittivity of free space = ε = 8.854×10⁻¹² F/m
Mass of proton m = 1.67x10⁻²⁷ kg
Required Information:
Energy radiated per second = dE/dt = ?
Answer:
[tex]\frac{dE}{dt} = 7.346x10^{14}[/tex] [tex]eV/s[/tex]
Explanation:
The rate at which energy is emitted from an accelerating charge with charge q and acceleration a is given by
[tex]\frac{dE}{dt} =\frac{q^{2}a^{2} }{6\pi{\epsilon}c^{3}}[/tex]
Where ε is the permittivity of free space and c is the speed of light
We know that centripetal acceleration is given by
[tex]a = \frac{v^2}{R}[/tex]
Whereas kinetic energy is given by
[tex]KE = \frac{1}{2}mv^{2}[/tex]
[tex]\frac{2KE}{m} = v^2[/tex]
Substitute in the equation of acceleration
[tex]a = \frac{2KE}{mR}[/tex]
[tex]a=1.13x10^{34}[/tex] [tex]eV/s^{2}[/tex]
Therefore, the corresponding energy radiated per second is
[tex]\frac{dE}{dt} =\frac{(1.60x10^{-19})^{2}(1.13x10^{34})^{2} }{6\pi{\epsilon}c^{3}}[/tex]
[tex]\frac{dE}{dt} =\frac{3.27x10^{30} }{6\pi8.854x10^{-12}(3x10^{8}^)^{3}}[/tex]
[tex]\frac{dE}{dt} = 7.346x10^{14}[/tex] [tex]eV/s[/tex]
The energy radiated by the proton is 7.346×10¹⁴ eV/s.
Electromagnetic radiation:
The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by:
[tex]\frac{dE}{dt}=\frac{q^2a^2}{6\pi\epsilon_oc^3}[/tex]
where q is the charge of the particle, and
a is the acceleration
It is given that a proton with kinetic energy (KE) of 5.0 MeV is traveling in a particle accelerator in a circular orbit with a radius (r) of 0.530 m
The acceleration of the particle in the circular orbit is given by:
[tex]a=\frac{v^2}{r}[/tex]
The kinetic energy is given by:
[tex]KE=\frac{1}{2}mv^2[/tex]
where m = 1.67×10⁻²⁷ kg is the mass of the proton
so,
[tex]5\times10^6\;eV=\frac{1}{2}\times1.67\times10^{-27}kg\times v^2\\\\v^2\approx6\times10^{33}eV/kg[/tex]
so acceleration:
[tex]a=\frac{v^2}{r}=\frac{6\times10^{33}}{0.53} \\\\a=1.13\times10^{34}\;eV/kgm[/tex]
now, we can calculate the energy radiated:
[tex]\frac{dE}{dt}=\frac{q^2a^2}{6\pi\epsilon_oc^3}=\frac{1.6\times10^{-19}\times(1.13\times10^{34})^2}{6\pi\times8.85\times10^{-12}\times(3\times10^8)^3}\\\\\frac{dE}{dt}=7.346\times10^{14}\;eV/s[/tex]
Learn more about electromagnetic radiations:
https://brainly.com/question/24470698?referrer=searchResults
https://brainly.com/question/17403491?referrer=searchResults
https://brainly.com/question/7596563?referrer=searchResults
