Question

A proton, a deuteron and an alpha particle enter a region of the magnetic field which is perpendicular to the velocity. If the kinetic energies are equal, then find the ratio of the radii.

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

Answer 1

The correct option is B

$\sqrt{2}:1:\sqrt{2}$

Step 1: Given data

The magnetic field is perpendicular to the velocity.

Kinetic energies are equal.

Step 2: To find

The ratio of radii of the proton, deuteron, and alpha particle

Step 3: Formula used

The radius of the circular path is given by,

$r=\frac{mv}{Bq}$

where $m$ is mass, $v$ is velocity, $B$ is the magnetic field and $q$ is charge on the particle

Linear momentum

$p=mv=\sqrt{2mK}$

where $K$ is the kinetic energy of the particle

Step 4: Finding the equation for radii in terms of mass and charge of the particle

From the above two equations, we get

$$\begin{gathered} r=\frac{\sqrt{2mK}}{Bq} \\ \Rightarrow r\propto \frac{\sqrt{m}}{q} \end{gathered}$$

Since $K$ and $B$ are constant

But we know,

charge and mass of the alpha particle,

$$\begin{gathered} q_{\alpha }=2q_p \\ {m}_{\alpha }=4m_p \end{gathered}$$

Here $q_p$= charge on proton

and $m_p$ = mass of proton.

charge and mass of deuteron,

$$\begin{gathered} q_d=2q_p \\ {m}_d=2m_p \end{gathered}$$

Step 5: Finding the ratio

So, the ratio of their radii,

$$\begin{gathered} r_p:r_d:r_{\alpha }=\frac{\sqrt{m_p}}{q_p}:\frac{\sqrt{m_d}}{q_d}:\frac{\sqrt{m_{\alpha }}}{q_{\alpha }} \\ \Rightarrow r_p:r_d:r_{\alpha }=\frac{\sqrt{m_p}}{q_p}:\frac{\sqrt{2m_p}}{2q_p}:\frac{\sqrt{4m_p}}{2q_p} \\ \Rightarrow r_p:r_d:r_{\alpha }=\sqrt{2}:1:\sqrt{2} \end{gathered}$$

The ratio of their radii is $\sqrt{2}:1:\sqrt{2}$

Option B is correct.