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

The Correct Option is C

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).