A proton and a deutron $(q=+\mathrm{e}, m=2.0 \mathrm{u})$ having 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 deutron path to the radius $r_p$ of the proton path is:

<p>To solve for the ratio of the radii of deutron and proton paths in a magnetic field, we use the formula for the radius $r$ of the circular path of a charged particle moving perpendicular to a uniform magnetic field:</p> <p>$r = \frac{mv}{qB}$</p> <p>where:</p> <ul> <li>$m$ is the mass of the particle,</li> <li>$v$ is the velocity of the particle,</li> <li>$q$ is the charge of the particle, and</li> <li>$B$ is the magnetic field strength.</li> </ul> <p>The proton and the deutron are given to have the same kinetic energy. The kinetic energy $K$ of a particle is given by:</p> <p>$K = \frac{1}{2}mv^2$</p> <p>From the kinetic energy, we can express the velocity as:</p> <p>$v = \sqrt{\frac{2K}{m}}$</p> <p>Since both particles have the same kinetic energy and the charge of the deutron is the same as the charge of the proton $q = e$, but the mass of the deutron is twice that of the proton ($m_d = 2m_p$), substituting the expression for $v$ in the radius formula, we get:</p> <p>For the deutron:</p> <p>$r_d = \frac{m_d\sqrt{2K/m_d}}{eB} = \sqrt{\frac{2K}{eB^2}} \cdot \sqrt{m_d}$</p> <p>For the proton ($m_p = m$):</p> <p>$r_p = \sqrt{\frac{2K}{eB^2}} \cdot \sqrt{m_p}$</p> <p>The ratio of the radius of the deutron path $r_d$ to the radius of the proton path $r_p$ is therefore:</p> <p>$$\frac{r_d}{r_p} = \frac{\sqrt{m_d}}{\sqrt{m_p}} = \sqrt{\frac{2m_p}{m_p}} = \sqrt{2}$$</p> <p>So, the correct answer is:</p> <p>Option C: $\sqrt{2}: 1$.</p>