Question

A proton, a deuteron, and an $\alpha$-particle accelerated through the same potential difference enter a region of uniform magnetic field, moving at right angles to $B$. What is the ratio of their kinetic energies?

Answer 1

Step 1: Given

A proton, a deuteron, and an $\alpha$-particle are accelerated through the same potential difference enter a region of uniform magnetic field and are moving at right angles to mangnetic field $B$.

As we know, the charge of a proton is $+1$.

A deuteron is the nucleus of the heavier isotope of hydrogen, deuterium (Element symbol ${}_1{}^2\mathrm{H}$ or $\mathrm{D}$). Thus, it is just a proton that is coupled with a neutron. Hence its charge is also $+1$.

An $\alpha$-particle is simply the nucleus of a helium-4 (${}_2{}^4\mathrm{He}$) atom. Thus, its charge is $+2$.

Let $V$ be the voltage under which the particles are accelerated.

Step 2: Formula used

The kinetic energy gained by a charged particle is
$K=qV$.
where $q$ is the charge of the particle and $V$ is the voltage under which it is accelerated.

Step 3: Calculate kinetic energy of a proton, deuteron and $\alpha$-particle

Kinetic energy of a proton, $K_p=+1\times V=V$

Kinetic energy of a deuteron, $K_d=+1\times V=V$

Kinetic energy of a $\alpha$-particle, $K_{\alpha }=+2\times V=2V$

Step 4: Calculate their ratio

$\begin{array}{cc}& K_p:K_d:K_{\alpha }=V:V:2V\\ \Rightarrow & K_p:K_d:K_{\alpha }=1:1:2\end{array}$

Therefore, the ratio of kinetic energies of a proton, deuteron, and $\alpha$-particle is $1:1:2$.