Question

A proton and a deutron both enter in a region uniform magnetic field B, moving at right angles to the field $\overrightarrow{B}$ . If the radius of circular orbits for both the particles is equal and kinetic energy acquired by deutron is I Me V, then kinetic energy acquired by the proton particle will be .

• A. 1 MeV
• B. 2 MeV
• C. 0.5 MeV
• D. 4 MeV

Answer 1

The correct option is A 1 MeV

$R=\frac{mv}{qB}=\frac{\sqrt{2mk}}{qB}{K}_p=\frac{q^2{B}^2{R}_p^2}{2mp}$

$k=\frac{q^2{B}^2{R}^2}{2m}$

$\therefore \frac{K_{\infty }}{K_p}={\left(\frac{q_{\infty }}{q{I}_0}\right)}^2\left(\frac{m_p}{m_{\infty }}\right){\left(\frac{R_p}{R_{\infty }}\right)}^2$

$\Rightarrow K_{\infty }=K_p{\left(\frac{q_{\infty }}{q_p}\right)}^2\left(\frac{mp}{m_{\infty }}\right){\left(\frac{R_{\infty }}{R_p}\right)}^2$

$K_p=1MeV{\left(2\right)}^2{\left(\frac{1}{4}\right)}^2{\left(1\right)}^2=1MeV$

Hence,

option $\left(A\right)$ is correct answer.