Respuesta :
Answer:
The value of fraction of energy is [tex]2.23\times10^{-11}[/tex]
The value of fraction of energy is [tex]7.53\times10^{-5}[/tex]
Explanation:
Given that,
Charge = q
Acceleration = a
The rate at which energy is emitted from an accelerating charge
[tex]\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon c^3}[/tex]....(I)
We know that,
Acceleration for circular motion is
[tex]a=\dfrac{v^2}{r}[/tex]....(II)
The kinetic energy is
[tex]K.E=\dfrac{1}{2}mv^2[/tex]
[tex]v^2=\dfrac{2(K.E)}{m}[/tex]
Put the value of v in equation (II)
[tex]a=\dfrac{2(K.E)}{mr}[/tex]
Put the value of a in equation (I)
[tex]\dfrac{dE}{dt}=\dfrac{q^2(\dfrac{2(K.E)}{mr})^2}{6\pi\epsilon c^3}[/tex]
[tex]\dfrac{dE}{dt}=\dfrac{q^24(K.E)^2}{6\pi\epsilon c^3\times m^2 r^2}[/tex]
[tex]\dfrac{\dfrac{dE}{dt}}{K.E}=\dfrac{q^24(K.E)}{6\pi\epsilon c^3\times m^2 r^2}[/tex]
Suppose that,
[tex]\dfrac{\dfrac{dE}{dt}}{K.E}=R[/tex]
So,
[tex]R=\dfrac{q^2\times4(K.E)}{6\pi\epsilon c^3\times m^2 r^2}[/tex]....(III)
(a). For proton,
We need to calculate the fraction of its energy does it radiate per second
Using equation (III)
[tex]R=\dfrac{4\times(1.6\times10^{-19})^2\times5.0\times1.6\times10^{-19}\times10^{6}\times6\times10^{9}}{(3\times10^{8})^3\times(1.67\times10^{-27})^2\times(0.540)^2}[/tex]
[tex]R=2.23\times10^{-11}[/tex]
(b). For electron,
We need to calculate the fraction of its energy does it radiate per second
Using equation (III)
[tex]R=\dfrac{4\times(1.6\times10^{-19})^2\times5.0\times1.6\times10^{-19}\times10^{6}\times6\times10^{9}}{(3\times10^{8})^3\times(9.1\times10^{-31})^2\times(0.540)^2}[/tex]
[tex]R=0.0000753[/tex]
Hence, The value of fraction of energy is [tex]2.23\times10^{-11}[/tex]
The value of fraction of energy is [tex]7.53\times10^{-5}[/tex]
