QJEEM24Ph31c5: JEE Main 2024 Physics - Class 12 JEEM24Ph31c5 Solution: A proton and a deutron having same kinetic energies enter a…

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:

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

where:

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

The proton and the deutron are given to have the same kinetic energy. The kinetic energy $K$ of a particle is given by:

$K = \frac{1}{2}mv^2$

From the kinetic energy, we can express the velocity as:

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

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:

For the deutron:

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

For the proton ( $m_p = m$ ):

$r_p = \sqrt{\frac{2K}{eB^2}} \cdot \sqrt{m_p}$

The ratio of the radius of the deutron path $r_d$ to the radius of the proton path $r_p$ is therefore:

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

So, the correct answer is:

Option C: $\sqrt{2}: 1$ .