Question

Three particles having charges in the ratio of 2:3:5 produce the same point on the photographic film in the Thomson experiment. Their masses are in the ratio of
(A) $2:3:5$
(B) $5:3:2$
(C) $15:10:6$
(D) $3:5:2$

Answer 1

Hint: In the question, the result of a Thomson experiment is given. J.J Thomson did this experiment to find the specific charge of electrons. In general Thomson, the experiment is used to find the specific charge of charged particles. We know the ratio of charges for the three particles. Using this ratio we have to find the ratio of the masses of the particles.

Complete step by step solution:
We know that, by definition
A specific charge is the ratio of charge of a particle to its mass.
The specific charge of any charged particle can be written as $\dfrac{q}{m}$
Where $q$ stands for the charge of the particle and $m$stands for the mass of the particle.
Since the three particles will produce the same point on the photographic film we can say that the three particles will have the same specific charge
i.e.
$\dfrac{q}{m} = constt.$
$\Rightarrow q \propto m$
From this, we can say that
$\dfrac{{{q_1}}}{{{m_1}}} = \dfrac{{{q_2}}}{{{m_2}}} = \dfrac{{{q_3}}}{{{m_3}}} = constt$
Since ${q_1}:{q_2}:{q_3} = 2:3:5$
The ratio of masses will also be the same in order to satisfy the above condition,
i.e.${m_1}:{m_2}:{m_3} = 2:3:5$

The answer is: Option (A): $2:3:5$

Note:
The main principle of the Thomson experiment is that when a beam of electrons is exposed to a magnetic field and an electric field it will experience a force. The net force experienced by the electron can be made zero by adjusting the magnitude and direction of the fields. The experiment is constructed using a discharge tube and two opposite charged electrodes. When we apply a potential difference across the two electrodes electrons will be emitted from the cathode. The force acting on these electrons is adjusted and the relation between the mass and charge is obtained.