Question

When the velocity of a relativistic charged particle increases, it’s specific charge
A. Decreases
B. Increases
C. Remains same
D. First decreases then increases

Answer 1

Hint: To answer this question, use the Einstein’s relativistic mass equation for a particle moving with speed v. Find the relation between mass and specific charge. Then, substitute this value in the formula for relativistic mass of the particle. Find the relation between specific charge and velocity of the relativistic charged particle. Use this relation to find the answer to this given question.

Formula used:
$m={\displaystyle \frac{m_0}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}}}$

Complete step by step answer:
Einstein’s relativistic mass equation for any particle moving with speed v is given by,
$m={\displaystyle \frac{m_0}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}}}$ …(1)
Where, m is the relativistic mass
             $m_0$ is the mass of the particle at rest
             c is the speed of the light
It is given that when the velocity of a relativistic charged particle increases, it means speed v increases.
Specific charge is given by,
$m={\displaystyle \frac{q}{m}}$ …(2)
Where, q is the charge
Substituting equation. (2) in equation. (1) we get,
${\displaystyle \frac{q}{m}}={\displaystyle \frac{{\displaystyle \frac{q}{m_0}}}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}}}$
From the above equation it can be inferred that,
${\displaystyle \frac{q}{m}}\propto {\displaystyle \frac{1}{v}}$
Thus, specific charge is inversely proportional to the velocity of the particle.
So, if the velocity of a relativistic charged particle increases, it’s specific charge decreases.
Hence, the correct answer is option B i.e. decreases.

Note:
Students must remember relativistic formulas to answer these types of questions. When velocity of a relativistic charged particle (v) is very less as compared to the speed of light i.e. c. Then, we consider ${\displaystyle \frac{v^2}{c^2}}=0$. But, when the speed of the particle is comparable to the speed of light then we cannot consider ${\displaystyle \frac{v^2}{c^2}}=0$.