Question

A particle of rest mass m0 is moving with speed v. Its classical kinetic energy is Kcl = 1 2m0v 2 . Let Krel be the relativistic expression for its kinetic energy. By expanding Krel/Kcl in powers of v 2/c2 , estimate the value of v 2/c2 for which Krel differs from Kcl by 10 percent. For this value of v 2/c2 , what is the kinetic energy in MeV of an electron and a proton? (Kleppner 13.3)

Answer 1

Given,

Rest of the particle $=m_o$

Speed of the particle $=v$

Classical kinetic energy $K_{cl }= \frac{m_ov^2 .}{2}$

Relativistic kinetic energy $K_{rel}=(1-\gamma) m_oc^2$

$\frac{K_{rel}-K_{cl}}{K_{rel}}\times 100=10\%$

$\Rightarrow \frac{m_oc^2\frac{-m_ov^2 .}{2}}{m_oc^2}=\frac{1}{10}$

$\Rightarrow \frac{c^2-\frac{v^2}{2}}{c^2}=\frac{1}{10}$

$\Rightarrow \frac{v^2}{2c^2}=\frac{9}{10}$

$\Rightarrow$

$K_{rel}=\frac{v^2}{2c^2}mc^2=\frac{1}{2}mv^2=K_{class}$

$K_{rel}=\frac{9}{10}mc^2=\frac{1}{2}mv^2=K_{class}$

For electron, $m=9.1\times 10^{-31}kg$

mass of proton (m)=$1.6\times 10^{-27}kg$

Now, substituting the values,

$K_{e}=\frac{9\times9.1\times 10^{-31}\times 3\times 10^8\times 3\times 10^8}{10}$

$=245.7\times 10^{-24}\times 3\times 10^8J$

$=2.45\times10^{-22}\times 3\times 10^8J$

$=0.4.5MeV$

For proton

$K_{p}=\frac{9\times 1.6\times 10^{-27}\times (3\times 10^8)^2}{10}$

$=\frac{129.6\times 10^{-12}}{1.6\times 10^{-13} } MeV$

$=810MeV$