Question

A proton and a deuteron ($q = +e, \, m = 2.0u$) having the same kinetic energies enter a region of uniform magnetic field $\vec{B}$, moving perpendicular to $\vec{B}$. The ratio of the radius $r_d$ of the deuteron path to the radius $r_p$ of the proton path is:

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

Answer 1

Answer: (C)

$\sqrt{2} : 1$

Answer 2

The radius of the path of a charged particle moving in a magnetic field is given by:

$r = \frac{mv}{qB}$,

where $m$ is the mass, $v$ is the velocity, $q$ is the charge, and $B$ is the magnetic field strength.

Since both particles have the same kinetic energy, $\frac{1}{2}mv^2 = K$, and the velocity $v$ can be expressed as:

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

Thus, the radius of the proton path $r_p$ is:

$r_p = \frac{m_p \sqrt{2K/m_p}}{eB}$,

and the radius of the deuteron path $r_d$ is:

$r_d = \frac{m_d \sqrt{2K/m_d}}{eB}$.

Since $m_d = 2m_p$, the ratio of the radii is:

$\frac{r_d}{r_p} = \frac{\sqrt{2m_p}}{\sqrt{m_p}} = \sqrt{2}$.

Thus, the ratio is $\sqrt{2} : 1$, and the correct answer is Option (3).