KCET · Physics · Ray Optics
A proton, a deuteron and an \( \alpha \) - particle are projected perpendicular to the direction of a
uniform magnetic field with same kinetic energy. The ratio of the radii of the circular paths
described by them is
- A \( 1: \sqrt{2}: 1 \)
- B \( 1: \sqrt{2}: \sqrt{2} \)
- C \( \sqrt{2}: 1: 1 \)
- D \( \sqrt{2}: \sqrt{2}: 1 \)
Answer & Solution
Correct Answer
(A) \( 1: \sqrt{2}: 1 \)
Step-by-step Solution
Detailed explanation
The radius of the circular path described by particle in uniform magnetic field is
\[
R=\frac{m v}{q B}=\frac{\sqrt{2 m E}}{q B}
\]
For proton, \( R_{p}=\frac{\sqrt{2 m_{p} E}}{q_{p} B} \)
For deuteron, \( R_{d}=\frac{\sqrt{2 m_{d} E}}{q_{d} B} \)
For a-particle, \( R_{\alpha}=\frac{\sqrt{2 m_{\alpha} E}}{q_{\alpha} B} \)
Now, \( q_{p}=e, q_{d}=e, q_{\alpha}=2 e \)
\( m_{p}=m, m_{d}=2 m, m_{\alpha}=4 m \)
Therefore,
\( R_{p}=\frac{\sqrt{2 m E}}{e B}, R_{d}=\frac{\sqrt{4 m E}}{e B}, R_{\alpha}=\frac{\sqrt{8 m E}}{2 e B}=\frac{\sqrt{2 m E}}{e B} \) \( \Rightarrow R_{p}: R_{d} ; R_{\alpha}=1: \sqrt{2}: 1 \)
\[
R=\frac{m v}{q B}=\frac{\sqrt{2 m E}}{q B}
\]
For proton, \( R_{p}=\frac{\sqrt{2 m_{p} E}}{q_{p} B} \)
For deuteron, \( R_{d}=\frac{\sqrt{2 m_{d} E}}{q_{d} B} \)
For a-particle, \( R_{\alpha}=\frac{\sqrt{2 m_{\alpha} E}}{q_{\alpha} B} \)
Now, \( q_{p}=e, q_{d}=e, q_{\alpha}=2 e \)
\( m_{p}=m, m_{d}=2 m, m_{\alpha}=4 m \)
Therefore,
\( R_{p}=\frac{\sqrt{2 m E}}{e B}, R_{d}=\frac{\sqrt{4 m E}}{e B}, R_{\alpha}=\frac{\sqrt{8 m E}}{2 e B}=\frac{\sqrt{2 m E}}{e B} \) \( \Rightarrow R_{p}: R_{d} ; R_{\alpha}=1: \sqrt{2}: 1 \)
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