Question

A proton with rest mass of 1.67 × 10^-27 kg moves in an accelerator with a speed of 0.6c. Its total energy is (c = 3 × 10⁸ m/s)
1. 1.88 × 10^-10 J
2. 5.46 × 10^-10 J
3. 0.92 × 10^-10 J
4. 3.42 × 10^-10 J

Answer 1

Correct Answer - Option 1 : 1.88 × 10^-10 J

Concept:

Variation of Mass with Velocity:

According to classical physics, the inertial mass of a body is independent of the velocity of light. It is regarding as a constant. However special theory of relativity leads us to the concept of variation of mass with velocity. It follows from the special theory of relativity that the mass m of a body moving with relativistic velocity v relative to an observer is larger than its m₀ when it is at rest.

According to Einstein, the mass of the body in motion is different from the mass of the body at rest.

\(m = \frac{{{m_o}}}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}\)

This is the relative formula for variation of mass with velocity where m₀ is the rest mass and m is the relativistic mass of the body.

Realistic Kinetic Energy:

The kinetic energy of a particle of rest mass m₀ where m is the mass of the particle when it is moving with a velocity c is:

\({E_k} = \left( {m - {m_o}} \right){c^2} \)

\(E_k= {m_o}{c^2}\left[ {\frac{1}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }} - 1} \right]\)

When v < c:

\({\left( {1 - \frac{{{v^2}}}{{{c^2}}}} \right)^{ - \frac{1}{2}}} \approx 1 + \frac{1}{2}\frac{{{v^2}}}{{{c^2}}}\)

\( \therefore {E_k} = \left( {m - {m_o}} \right){c^2} \)

\(= {m_o}{c^2}\left( {\frac{1}{2}\frac{{{v^2}}}{{{c^2}}}} \right) = \frac{1}{2}{m_o}{v^2}\)

Total energy will be:

\(E = m{c^2} = \frac{{{m_o}{c^2}}}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}\)   ---(1)

Calculation:

m₀ = 1.67 × 10^-27 kg, v = 0.6 c, c = 3 × 10⁸ m/s

Putting on the respective values in Equation (1), we get:

\({E_k} = \frac{{1.67 \times {{10}^{ - 27}} \times {{\left( {3 \times {{10}^8}} \right)}^2}}}{{\sqrt {1 - {{\left( {0.6} \right)}^2}} }}\)

\( E_k= 1.88 \times {10^{ - 10}}\;J\)