Question

The specific charge of a proton is 9.6$ \times {10^7}$ C/Kg. The specific charge of an alpha particle will be:
$
  (a){\text{ 9}}{\text{.6}} \times {\text{1}}{{\text{0}}^7}C/kg \\
  (b){\text{ 19}}{\text{.2}} \times {\text{1}}{{\text{0}}^7}C/kg \\
  (c){\text{ 4}}{\text{.8}} \times {\text{1}}{{\text{0}}^7}C/kg \\
  (d){\text{ 2}}{\text{.4}} \times {\text{1}}{{\text{0}}^7}C/kg \\
$

Answer 1

Hint:In this question use the concept that the specific charge is simply the ratio of charge (q) to mass (m). The charge of alpha particle is 2q that is twice the charge on proton whereas mass of alpha particle is 4m where m is the mass of proton. This will help approaching the problem.


Complete step-by-step solution -

Given data:
Specific charge of a proton = 9.6$ \times {10^7}$ C/Kg.
Now we have to find out the specific charge on an alpha particle.
As we know specific charge is the ratio of charge (q) of proton to mass (m) of proton
Therefore, (q/m) = 9.6$ \times {10^7}$ C/Kg........................ (1), (specific charge of proton)
Now as we know that charge on an alpha particle will be twice the proton and mass of the alpha particle will be four times the proton.
Let charge and mass of the alpha particle be q’ and m’ respectively.
Therefore charge on the alpha particle, q’ = 2q.
And mass on the alpha particle, m’ = 4m.
So the specific charge on the alpha particle is (q’/m’).
Therefore,
Specific charge on the alpha particle, (q’/m’) = (2q/4m) = $\dfrac{1}{2}$(q/m)
Now from equation (1) we have,
Specific charge on the alpha particle, (q’/m’) = $\dfrac{1}{2}$(9.6$ \times {10^7}$) = 4.8$ \times {10^7}$ C/Kg.
So this is the required specific charge on the alpha particle.
Hence option (C) is the correct answer.

Note – Alpha particles are somewhat similar to a helium-4 nucleus. The main reason behind its charge being twice the charge of a proton is that it is composed of two protons and two neutrons which are bound together. The formation of alpha particles is generally accompanied due to alpha decay of a radioactive nucleus.