The speed of a fast moving electron, having total energy of 2 MeV, is nearly:

Concept:

A relativistic particle is a particle that moves with a relativistic speed; that is, a speed comparable to the speed of light.

Here the speed is comparable to the light, hence we should follow the relativistic approach

In the relativistic approach, the total energy is given by:

E = rest energy + K.E

\(E = m{c^2} + \left( {\gamma - 1} \right)m{c^2} = \gamma m{c^2}\)

Where \(\gamma = \frac{1}{{\sqrt {1 - {{\left( {\frac{v}{c}} \right)}^2}} }}\)

Calculation:

Given particle is electron, the mass of electron is m = 9.11 × 10^-31 kg, c = 3 × 10⁸ m/s

1 eV = 1.602 × 10-19 J

E = 2 MeV = 2 × 10⁶ × 1.602 × 10^-19 J = 3.204 × 10^-13 J 

Now from the above formula:

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

\( 3.204 \times {10^{ - 13}} = \frac{1}{{\sqrt {1 - {{\left( {\frac{v}{c}} \right)}^2}} }} \times 9.11 \times {10^{ - 31}} \times {\left( {3 \times {{10}^8}} \right)^2}\)

\(\frac{1}{{\sqrt {1 - {{\left( {\frac{v}{c}} \right)}^2}} }} = 3.9078\)

\(\frac{v}{c} = 0.9667\)

v = 0.97 c

If we take directly K.E = 0.5 mv²

\(3.204 \times {10^{ - 13}} = 0.5 \times 9.11 \times {10^{ - 31}} \times {v^2} \Rightarrow v = 8.38 \times {10^8}\;m/s\)

Which is way more than the speed of light, so not possible.

Note: Mass of proton = 1.67 × 10^-27 kg, Mass of Neutron = 1.67 × 10^-27 kg;