Question

The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by dE/dt= q^2 a^2/ 6πεc^3, where c is the speed of light.

Required:
a. If a proton with a kinetic energy of 5.0 MeV is travelling in a particle accelerator in a circular orbit with a radius of 0.540 m, what fraction of its energy does it radiate per second?
b. Consider an electron orbiting with the same speed and radius. What fraction of its energy does it radiate per second?

Answer 1

The value of fraction of energy is
`2.23*10^(-11)`

The value of fraction of energy is
`7.53*10^(-5)`

Step-by-step explanation:

Given that,

Charge = q

Acceleration = a

The rate at which energy is emitted from an accelerating charge

`(dE)/(dt)=(q^2a^2)/(6\pi\epsilon c^3)`....(I)

We know that,

Acceleration for circular motion is

`a=(v^2)/(r)`....(II)

The kinetic energy is

`K.E=(1)/(2)mv^2`

`v^2=(2(K.E))/(m)`

Put the value of v in equation (II)

`a=(2(K.E))/(mr)`

Put the value of a in equation (I)

`(dE)/(dt)=(q^2((2(K.E))/(mr))^2)/(6\pi\epsilon c^3)`

`(dE)/(dt)=(q^24(K.E)^2)/(6\pi\epsilon c^3* m^2 r^2)`

`((dE)/(dt))/(K.E)=(q^24(K.E))/(6\pi\epsilon c^3* m^2 r^2)`

Suppose that,

`((dE)/(dt))/(K.E)=R`

So,

`R=(q^2*4(K.E))/(6\pi\epsilon c^3* m^2 r^2)`....(III)

(a). For proton,

We need to calculate the fraction of its energy does it radiate per second

Using equation (III)

`R=(4*(1.6*10^(-19))^2*5.0*1.6*10^(-19)*10^(6)*6*10^(9))/((3*10^(8))^3*(1.67*10^(-27))^2*(0.540)^2)`

`R=2.23*10^(-11)`

(b). For electron,

We need to calculate the fraction of its energy does it radiate per second

Using equation (III)

`R=(4*(1.6*10^(-19))^2*5.0*1.6*10^(-19)*10^(6)*6*10^(9))/((3*10^(8))^3*(9.1*10^(-31))^2*(0.540)^2)`

`R=0.0000753`

Hence, The value of fraction of energy is
`2.23*10^(-11)`

The value of fraction of energy is
`7.53*10^(-5)`