Q143 / 200 · Manipal (MET) 2011 · Physics
A proton and an $\alpha$-particle enter a uniform magnetic field perpendicularly with the same speed. If proton takes $25 \mu \mathrm{~s}$ to make 5 revolutions, then the periodic time for the $\alpha$-particle would be
- A$50 \mu \mathrm{~s}$
- B$25 \mu \mathrm{~s}$
- Correct answerC$10 \mu \mathrm{~s}$
- D$5 \mu \mathrm{~s}$
Solution
Time period of proton $T_p=\frac{25}{5}=5 \mu \mathrm{~s}$
By using $T=\frac{2 \pi m}{q B}$
$\begin{aligned} & \Rightarrow \quad \frac{T_\alpha}{T_p}=\frac{m_\alpha}{m_p} \times \frac{q_p}{q_\alpha}=\frac{4 m_p}{m_p} \times \frac{q_p}{2 q_p}=2 \\ & \Rightarrow T_\alpha=2 T_p=10 \mu \mathrm{~s}\end{aligned}$
Hence, the correct answer is option: 'C'.