Question

Deuteron and alpha particles having the same KE enter in a magnetic field. If the ratio of the radius of Deuteron and alpha particle is $x\sqrt{2}$ . Then $x=$

• A. $5$
• B. $8$
• C. $3$
• D. $1$

Answer 1

The correct option is D

$1$

Step 1. The radius of the charged particle, $r$

$r=\frac{mv}{qB}$ Where, [$m$ is the mass of the charged particle]

[$v$ is the velocity of the charged particle]

[$q$ is the charge]

[$B$ is the magnetic field]

Step 2. The radius of the charged particle in terms of the kinetic energy, $K.E.$

We know that, the formula of Kinetic energy is

$K.E.=\frac{1}{2}\times m\times v^2$

$v=\sqrt{\frac{2\times K.E.}{m}}$

From the formula of the radius of the charged particle,

$r=\frac{m}{qB}\sqrt{\frac{2\times K.E.}{m}}$

$r=\frac{\sqrt{2\times m\times K.E.}}{q\times B}$

Step 3. Finding the ratio of the radius of the deuteron and the alpha particle.

As $K.E.$ and $B$ are same then, from the above formula

$\frac{r_d}{r_{\alpha }}=\frac{{\displaystyle \frac{\sqrt{2\times m_d\times K.E.}}{q_d\times B}}}{{\displaystyle \frac{\sqrt{2\times m_{\alpha }\times K.E.}}{q_{\alpha }\times B}}}$ where, [$m_d$ is the mass of deuteron]

[$m_{\alpha }$ is the mass of alpha particles]

[$q_d,q_{\alpha }$ are the charges on deuteron and alpha particles respectively]

$\frac{r_d}{r_{\alpha }}=\sqrt{\frac{m_d}{m_{\alpha }}}\times \frac{q_{\alpha }}{q_d}$ [ $m_d=2\times$mass of proton $m_p$ and $m_{\alpha }=4\times$mass of proton]

[$q_{\alpha }=2\times q_d$]

$\frac{r_d}{r_{\alpha }}=\sqrt{\frac{2\times m_p}{4\times m_p}}\frac{2\times q_d}{q_d}$

$\frac{r_d}{r_{\alpha }}=\sqrt{\frac{2}{4}}\times 2$

$\frac{r_{{}_d}}{r_{\alpha }}=\sqrt{\frac{1}{2}}\times 2$

$\frac{r_d}{r_{\alpha }}=\frac{2\times \sqrt{2}}{\sqrt{2}\times \sqrt{2}}$ [By rationalizing]

$\frac{r_d}{r_{\alpha }}=\sqrt{2}$

Therefore, $x=1$

Hence, the correct option is $\left(D\right)$.