Question

When the speed of electron increases, the specific charge:
(A) decreases
(B) increases
(C) remains the same
(D) none of the above

Answer 1

Hint: The specific charge of a species is defined as the ratio of the charge of the species to its mass, and in turn, the mass of a species depends upon the speed with which it moves. So, eventually, the variation in specific charge of an electron can be determined by first establishing a relationship between specific charge and mass, and then between mass and speed of the electron.

Complete step by step answer:
The specific charge of an electron is given by the following formula:

$Specific\; charge = \dfrac{charge}{mass} …(i)$

The charge of an electron is fixed. So, the specific charge of an electron varies only with its mass. The mass of a moving electron is given by the following relation:

$mass = \dfrac{m_e}{\sqrt{1-{\dfrac{u}{c}}^2}}$

Or, $mass = \dfrac{m_e}{\sqrt{\dfrac{c^2 – u^2}{c^2}}}$ …$(ii)$

Where, $m_e$ is the mass of the electron at rest;
$u$ is the speed or velocity of the moving electron;
$c$ is the velocity of light $3\times 10^8\ m/s$.
From the above relation, we observe that the mass of an electron is directly proportional to the speed of the electron, which means that if the speed of an electron increases, its mass will also increase.
Also, from the relation in equation $(i)$, it can be deduced that the specific charge of an electron is inversely proportional to its mass.
Taking both the equations into account, we find that if the speed of an electron increases, its mass increases too, and if the mass increases, the specific charge of the electron decreases. So, combining both the relationships, it can be established that when the speed of the electron increases, the specific charge decreases.

So, the correct answer is Option A .

Note: If velocity increases, then the value \[{c^2} - {v^2}\] will decrease. So, the mass will eventually increase. The charge of an electron is fixed, which is, \[ - 1.6 \times {10^{ - 19}}coulombs\]. When the speed of the electron increases, the De Broglie wavelength of the electron decreases.