Question

An alpha particle and a proton are accelerated from rest in a uniform electric field. The ratio of the times taken by proton and alpha particle to attain equal displacements is

• A. $\sqrt{2}:1$
• B. $1:2$
• C. $1:\sqrt{2}$
• D. $2:1$

Hint:

For problems comparing the motion of different charged particles in an electric field, first derive a general expression for the quantity of interest (like time, velocity, or kinetic energy) in terms of mass ($m$) and charge ($q$). Then, use the proportionality to set up a ratio. Remember that an alpha particle is a helium nucleus ($^4_2$He), with mass $\approx 4m_p$ and charge $+2e$.

Answer 1

The Correct Option is C

Let a particle of charge $q$ and mass $m$ be accelerated from rest in a uniform electric field $E$.
The force on the particle is $F = qE$.
The acceleration of the particle is $a = \frac{F}{m} = \frac{qE}{m}$.
The particle starts from rest ($u=0$) and covers a displacement $s$. We can use the kinematic equation $s = ut + \frac{1}{2}at^2$.
$s = 0 + \frac{1}{2}at^2 = \frac{1}{2} \left(\frac{qE}{m}\right)t^2$.
We can solve this for the time taken, $t$:
$t^2 = \frac{2sm}{qE} \implies t = \sqrt{\frac{2sm}{qE}}$.
Since the displacement $s$ and the electric field $E$ are the same for both particles, we can say that $t \propto \sqrt{\frac{m}{q}}$.
Let's find the mass and charge for the proton and alpha particle.
For a proton (p): Mass $m_p$, Charge $q_p = e$.
For an alpha particle ($\alpha$): Mass $m_\alpha \approx 4m_p$, Charge $q_\alpha = 2e$.
Now we can find the ratio of the times, $\frac{t_p}{t_\alpha}$.
$\frac{t_p}{t_\alpha} = \frac{\sqrt{m_p/q_p}}{\sqrt{m_\alpha/q_\alpha}} = \sqrt{\frac{m_p}{q_p} \cdot \frac{q_\alpha}{m_\alpha}}$.
$\frac{t_p}{t_\alpha} = \sqrt{\frac{m_p}{e} \cdot \frac{2e}{4m_p}} = \sqrt{\frac{2}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
The ratio of the times taken by the proton and the alpha particle is $1:\sqrt{2}$.