Question

The factor $\gamma$appears in many relativistic expressions. A value $\gamma =1.01$implies that relativity changes the Newtonian values by approximately $1\%$and that relativistic effects can no longer be ignored. At what kinetic energy, in $\mathrm{MeV}$, is $\gamma =1.01$for

(a) an electron,

(b) a proton, and

(c) an alpha particle?

Short answer

(a) The kinetic energy for an electron is $E_k=5.12\times {10}^3\mathrm{eV}$

(b) The kinetic energy for a proton is $E_k==9.39\times {10}^6\mathrm{eV}$

(c) The kinetic energy for an alpha particle is $E_k=3.73\times {10}^7\mathrm{eV}$

Step-by-step solution

Given Information

Kinetic energy according to relativity is given by $\gamma m{c}^2-m{c}^2=m{c}^2(\gamma -1)=0.01m{c}^2$
Where $c$ is the velocity of light in vacuum and is equal to .$c=3\times {10}^8$

Part (a) Step 1 Calculation

For electron, $m=9.1\times {10}^{-31}$kg. So kinetic energy is,

$E_k=0.01\times 9.1\times {10}^{-31}\times {\left(3\times {10}^8\right)}^2=8.129\times {10}^{-16}\mathrm{J}=5.12\times {10}^3\mathrm{eV}$

Part (b) Step 1 Calculation

Mass of the proton is, $m=1.67\times {10}^{-27}$kg. So, kinetic energy is,

$E_k=0.1\left(1.67\times {10}^{-27}\times {\left(3\times {10}^8\right)}^2\right)=1.50\times {10}^{-12}\mathrm{J}=9.39\times {10}^6\mathrm{eV}$

Part (c) Step 1 Calculation

The mass of the alpha particle is, $m=6.64\times {10}^{-27}$kg. So the kinetic energy is,

$E_k=0.01\times 6.64\times {10}^{-27}\times {\left(3\times {10}^8\right)}^2=5.98\times {10}^{-12}\mathrm{J}=3.73\times {10}^7\mathrm{eV}$