Question

What is the e/m ratio of an electron?

Answer 1

Hint: J.J Thomson measures the e/m ratio of an electron. e/m, the ratio of charge of an electron to the mass of an electron. Here m is the mass of the particle of cathode ray in kg and e is its charge in coulomb.

Complete step by step solution:
In 1897 J.J Thomson first measured the e/m ratio of an electron.
- He said when an electron enters a region in which there is a uniform magnetic field, B, perpendicular to the velocity, v, of the electron he electron experiences a force, F which is equal to $$F=e\times v\times B$$
The force is perpendicular to both v and B and its direction can be found by using the right-hand rule. This force will cause the electron to move in a circular orbit with radius r. According to Newton's law of Newton these two forces should be equal. So equating these two forces we get:$\text{e × v × B=m}{\displaystyle \frac{{\text{v}}^{\text{2}}}{\text{r}}}$
solving this equation we get ${\displaystyle \frac{\text{e}}{\text{m}}}\text{=}{\displaystyle \frac{\text{v}}{\text{Br}}}$
 Now all we need is the velocity of the electron which can be calculated by the formula:
${\displaystyle \frac{\text{1}}{\text{2}}}\text{m}{\text{v}}^{\text{2}}\text{=e}{\text{V}}_{\text{acc}}$
Solving the value of v from above step we get as:$\text{v=}\sqrt{\text{2e}{\text{V}}_{\text{acc}}}÷\sqrt{\text{m}}$
Now substituting this value of v in step 3 we get as ${\displaystyle \frac{\text{e}}{\text{m}}}\text{=}{\displaystyle \frac{\text{2}{\text{V}}_{\text{acc}}}{{\text{B}}^{\text{2}}{\text{r}}^{\text{2}}}}$ where ${\text{V}}_{\text{acc}}$ is the potential difference, B is the magnetic field and r is the radius of the electron's circuit orbit
So from above, we can say that e/m ratio of an electron is equal to $$\text{1}\text{.758820 }\times \text{ 1}{\text{0}}^{\text{11}}\text{Ck}{\text{g}}^{\text{-1}}$$

Note: The value of e/m remained constant irrespective of the nature of the gas. It means whether you use helium gas, neon gas, carbon gas and whatsoever the value of e/m ratio will remain constant.