Q51CP - University Physics with Modern Physics

Question

Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by $\frac{\mathbf{dE}}{\mathbf{dt}}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}{\mathbf{a}}^{\mathbf{2}}}{\mathbf{6}{\mathbf{\tau \tau \epsilon }}_{\mathbf{0}}{\mathbf{c}}^{\mathbf{3}}}$ where c is the speed of light. (a) Verify that this equation is dimensionally correct. (b) If a proton with a kinetic energy of 6.0 MeV is traveling in a particle accelerator in a circular orbit of radius 0.750 m, what fraction of its energy does it radiate per second? (c) Consider an electron orbiting with the same speed and radius. What fraction of its energy does it radiate per second?

Step-by-Step Solution

Answer

A) The rate of energy emission from an accelerating charge $\frac{\mathrm{dE}}{\mathrm{dt}} \mathrm{is} \frac{{\mathrm{q}}^2{\mathrm{a}}^2}{6{\mathrm{\pi \epsilon }}_0{\mathrm{c}}^3}$ and it is verified that dimensionally correct B) The fraction of energy radiates per second is $1.39\times {10}^{-11} \mathrm{persec}$ C) The fraction of energy radiates per second electron orbiting with the same speed and radius is $2.54\times {10}^{-8}$ 

STEP 1 Concept of the rate at which energy is emitted from an acceleration charge and the acceleration of the circular orbit

The rate at which energy is emitted from an acceleration charge q and the acceleration a is $\frac{\mathbf{dE}}{\mathbf{dt}}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}{\mathbf{a}}^{\mathbf{2}}}{\mathbf{6}{\mathbf{\tau \tau \epsilon }}_{\mathbf{0}}{\mathbf{c}}^{\mathbf{3}}}$ q is the acceleration charge, a is the acceleration of the hydrogen atom, ${\mathbf{\epsilon }}_{\mathbf{0}}$ is the electric constant, c is the speed of the light, The acceleration of the circular orbit is given as $\mathbf{a}\mathbf{=}\frac{{\mathbf{v}}^{\mathbf{2}}}{\mathbf{R}}$v is the velocity of the circular orbit, R is the radius of the circular orbit

STEP 2 To verify the equation

The rate at which energy is emitted from an acceleration charge that has charge q and acceleration a is given by $\frac{\mathrm{dE}}{\mathrm{dt}}=\frac{{\mathrm{q}}^2{\mathrm{a}}^2}{6{\mathrm{\tau \tau \epsilon }}_0{\mathrm{c}}^3}$ Substitute the units we have,

$$\begin{gathered} \frac{\mathrm{dE}}{\mathrm{dt}}=\frac{{\mathrm{C}}^2}{{\mathrm{C}}^2\left(\mathrm{N}-{\mathrm{m}}^2\right)\left(\mathrm{m}/{\mathrm{s}}^2\right)} \\ =\mathrm{J}/\mathrm{s} \\ \frac{\mathrm{dE}}{\mathrm{dt}}=\mathrm{W} \end{gathered}$$

Thus, the rate of energy emission from an accelerating charge $\frac{\mathrm{dE}}{\mathrm{dt}} \mathrm{is} \frac{{\mathrm{q}}^2{\mathrm{a}}^2}{6{\mathrm{\epsilon }}_0{\mathrm{c}}^2}$and it is verified that dimensionally correct

STEP 3: Calculate the acceleration of the circular orbit

The acceleration of the circular orbit is $\mathrm{a}=\frac{{\mathrm{v}}^2}{\mathrm{R}}$ Multiply numerator and denominator with $\frac{1}{2}\mathrm{m}$ we get,$\mathrm{a}=\frac{{\displaystyle \frac{1}{2}}{\mathrm{mv}}^2}{\frac{1}{2}\mathrm{mR}}$ Kinetic energy of the proton is ${\mathrm{K}}_{\mathrm{P}}=6.0\times {10}^6\mathrm{eV}$. Substitute the values we have,

$$\begin{gathered} {\mathrm{K}}_{\mathrm{p}}=6.0\times {10}^6\mathrm{eV}\times 1.6\times {10}^{-19}\mathrm{J}/\mathrm{ev} \\ =9.6\times {10}^{-13} \mathrm{J} \end{gathered}$$

Substitute ${\mathrm{K}}_{\mathrm{p}}$ as $\frac{1}{2}\mathrm{m}$ in $\mathrm{a}=\frac{{\mathrm{v}}^2}{\mathrm{R}}$ we get,

$$\begin{gathered} \mathrm{a}=\frac{{\mathrm{K}}_{\mathrm{P}}}{{\displaystyle \frac{1}{2}}\mathrm{mR}} \\ =\frac{2{\mathrm{K}}_{\mathrm{P}}}{\mathrm{mR}} \\ \mathrm{a}=\frac{2\left(9.6\times {10}^{-13} \mathrm{J}\right)}{\left(1.67\times {10}^{-27}\mathrm{kg}\right)\left(0.750\mathrm{m}\right)} \\ =1.53\times {10}^{15}\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

 

Thus, the acceleration of the circular orbit is $1.53\times {10}^{15}\mathrm{m}/{\mathrm{s}}^2$

STEP 4: Calculate the energy required for the proton

The rate at which energy is emitted from an acceleration charge q and the acceleration a is $\frac{\mathrm{dE}}{\mathrm{dt}}=\frac{{\mathrm{q}}^2{\mathrm{a}}^2}{6{\mathrm{\pi \epsilon }}_0{\mathrm{c}}^3}$ Substitute the values we get,

$$\begin{gathered} \frac{\mathrm{dE}}{\mathrm{dt}}=\frac{{\left(1.60\times {10}^{-19}\right)}^{2}1.53\times {10}^{15}}{6\mathrm{\pi }\left(8.854\times {10}^{-12}\right)\left(2.998\times {10}^8\right)} \\ =1.33\times {10}^{-23}\mathrm{Nm}/\mathrm{s} \\ =1.33\times {10}^{-23}\mathrm{J}/\mathrm{s} \end{gathered}$$

To convert energy in electron volt into energy in joules is $1.53\times {10}^{15}\mathrm{m}/{\mathrm{s}}^2$ 

$$\begin{gathered} {\mathrm{E}}_{\mathrm{ev}}=1.33\times {10}^{-23}\mathrm{J}/\mathrm{s}\left(6.241\times {10}^{18}\right) \\ =8.32\times {10}^{-5}\mathrm{ev}/\mathrm{s} \end{gathered}$$

The fraction of its energy radiates per second is 

$$\begin{gathered} \frac{{\displaystyle \frac{\mathrm{dE}}{\mathrm{dt}}}\left(1\mathrm{s}\right)}{\mathrm{E}}=\frac{{\mathrm{E}}_{\mathrm{proton}}}{{\mathrm{K}}_{\mathrm{p}}} \\ \frac{{\displaystyle \frac{\mathrm{dE}}{\mathrm{dt}}\left(1\mathrm{s}\right)}}{\mathrm{E}}=\frac{8.32\times {10}^{-5}\mathrm{ev}/\mathrm{s}}{6.0\times {10}^6\mathrm{eV}} \\ =1.39\times {10}^{-11}\mathrm{per} \sec \end{gathered}$$

Thus, the fraction of energy radiates per second is $1.39\times {10}^{-11}\mathrm{per} \sec$

STEP 5:Calculate the fraction of energy radiate per second electron orbiting with the same speed and radius

The rate at which emits the energy of the electron is $\frac{\mathrm{dE}}{\mathrm{dt}}=8.32\times {10}^{-5} \mathrm{ev}/\mathrm{s}$ Initial energy of the proton is differs from the electron by the ratio of their masses to find the energy of the electron is ${\mathrm{E}}_0={\mathrm{E}}_{\mathrm{p}}\left(\frac{{\mathrm{m}}_{\mathrm{e}}}{{\mathrm{m}}_{\mathrm{p}}}\right)$ Ep is the energy of the electron, me is the mass of the electron, mp is the mass of the proton. Substitute the values we have,

$$\begin{gathered} {\mathrm{E}}_0=6.0\times {10}^6\mathrm{eV}\left(\frac{9.11\times {10}^{-11}\mathrm{kg}}{\left(1.67\times {10}^{-27} \mathrm{kg}\right)}\right) \\ =3273\mathrm{eV} \end{gathered}$$

Thus, the initial energy of the electron is 3273 eV. 

The fraction of energy radiates per second is

$$\begin{gathered} \frac{{\displaystyle \frac{dE}{dt}}\left(1s\right)}{E_e}=\frac{8.32\times {10}^{-5}eV/s\left(1s\right)}{3273eV} \\ =2.54\times {10}^{-8} \end{gathered}$$

Therefore, the fraction of energy radiate per second is $2.54\times {10}^{-8}$