Question

A proton ( m=1.67 × 10 -27 kg) is accelerated to a kinetic energy of 200 MeV. What is its speed at this energy?

A-1.70 × 108 m/s

B-9.79 × 107 m/s

C-2.47 × 108 m/s

D-6.25 × 108 m/s

Answer 1

Answer 2

Given:

• Mass of proton, $m=1.67\times 10^{-27}\text{kg}$

• Kinetic energy, $KE=200\text{MeV}=200\times 1.602\times 10^{-13}\text{J}=3.204\times 10^{-11}\text{J}$

• Rest mass energy of proton, $E_0=m{c}^2=938\text{MeV}$

Relativistic Kinetic Energy Formula:

$$KE=(\gamma -1)E_0$$

Where:

• $\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ (Lorentz factor)

• $c$ is the speed of light ($3\times 10^8\text{m/s}$)

• 
  

Step 1: Solve for $\gamma$$\gamma$

$$\gamma -1=\frac{KE}{E_0}$$$$\gamma =1+\frac{200\text{MeV}}{938\text{MeV}}$$$$\gamma \approx 1+0.213=1.213$$

Step 2: Solve for $v$$v$

$$\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$$$\sqrt{1-\frac{v^2}{c^2}}=\frac{1}{\gamma }$$$$1-\frac{v^2}{c^2}={\left(\frac{1}{\gamma }\right)}^2$$$$\frac{v^2}{c^2}=1-{\left(\frac{1}{1.213}\right)}^2$$$$\frac{v^2}{c^2}\approx 1-0.680=0.320$$$$v=c\sqrt{0.320}$$$$v\approx 3\times 10^8\times 0.566$$$$v\approx 1.70\times 10^8\text{m/s}$$

Answer:

The speed of the proton is A-1.70 × 10⁸ m/s