Question

The ratio of the $\dfrac{e}{m}$ (specific charge) values of an electron and an $\alpha -$ particle is –
A. $2:1$
B. $1:1$
C. $1:2$
D. none of these

Answer 1

Specific charge is the ratio between charge possessed by a substance and its mass. An $\alpha -$ particle is a doubly charged helium atom, consisting of no electrons. Hence, we must first find a relation between the masses of the alpha particle and the electron. We can then find the specific charge of each particle, and divide them to get the ratio.

Formulas used: Specific charge $= \dfrac{e}{m}$
Where $e$ is the charge of the particle and $m$ is the mass of the particle.

Complete step by step answer:
Since the specific charge is the charge per unit mass, it can be given as:
Specific charge $= \dfrac{e}{m}$
Where $e$ is the charge of the particle and $m$ is the mass of the particle.
An electron has a charge which can be denoted by “$e$”.
The mass of an electron can be denoted by ${m_e} = 9.1 \times {10^{ - 31}}kg$
Hence, the specific charge of an electron $= \dfrac{e}{{{m_e}}}$……………… (1)
An alpha particle is a doubly charged helium atom ($H{e^{2 + }}$). Thus, it has no electrons and only a nucleus. As we know, helium has two protons and two neutrons, making its mass number $= 4$. Since the mass of a proton and neutron are the same, the mass of the $\alpha -$ particle is:
${M_\alpha } = 4{m_p}$
Where ${M_\alpha }$ is the mass of the alpha particle and ${m_p}$ is the mass of a proton.
The mass of a proton is ${m_p} = 1.6724 \times {10^{ - 27}}kg$
Dividing this with the mass of an electron, ${m_e} = 9.1 \times {10^{ - 31}}kg$, we get:
$\dfrac{{{m_p}}}{{{m_e}}} = \dfrac{{1.6724 \times {{10}^{ - 27}}kg}}{{9.1 \times {{10}^{ - 31}}kg}}$
On solving this, we get:
$\dfrac{{{m_p}}}{{{m_e}}} = 1836$
$\Rightarrow {m_p} = 1836{m_e}$
Substituting this in our equation for the mass of an $\alpha -$ particle, we get:
${M_\alpha } = 4 \times 1836 \times {m_e} = 7344{m_e}$
And since the $\alpha -$ particle ($H{e^{2 + }}$) has a double positive charge on it, its charge can be denoted as “$2e$”
Hence, the specific charge of an $\alpha -$ particle $= \dfrac{{2e}}{{{M_\alpha }}}$
Substituting the value of ${M_\alpha }$ we got earlier, we get:
Specific charge of $\alpha -$ particle $= \dfrac{{2e}}{{7344{m_e}}} = \dfrac{e}{{3672{m_e}}}$………………… (2)
Dividing equation (1) with equation (2) will give us our answer:
Ratio of specific charges of an electron to an $\alpha -$ particle
$= \dfrac{{\dfrac{e}{{{m_e}}}}}{{\dfrac{e}{{3672{m_e}}}}}$
On solving this, we get:
Ratio of specific charges of an electron to an $\alpha -$ particle = 3672:1

Since this answer is not mentioned in the options, the correct option to be marked is D (none of these).

Note: When considering the charge of the two particles, we neglect the fact that one is negative while the other is positive. This is because we are considering the ratio only, which is mentioned in absolute terms. Note that an $\alpha -$ particle is a highly energetic particle produced in radioactive decay. J.J Thompson was the first scientist to demonstrate the specific charge of an electron.