When a moving proton is subjected to a uniform magnetic field it produces 5 revolutions in \( 25\,\mu\mathrm{s} \). What will be the time period of revolution for a moving \( \alpha \)-particle subjected to the same magnetic field.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 10\,\mu\mathrm{s} \)
A charged particle moving perpendicular to a uniform field moves in a circle with a time period independent of its speed:
\( \displaystyle T = \frac{2\pi m}{q B} \;\Rightarrow\; T \propto \frac{m}{q} \)
For the proton, 5 revolutions take \( 25\,\mu\mathrm{s} \):
\( \displaystyle T_p = \frac{25\,\mu\mathrm{s}}{5} = 5\,\mu\mathrm{s} \)
An alpha particle has \( m_\alpha = 4m_p \) and \( q_\alpha = 2e \):
\( \displaystyle \frac{T_\alpha}{T_p} = \frac{m_\alpha/q_\alpha}{m_p/q_p} = \frac{4m_p/2e}{m_p/e} = 2 \)
\( \displaystyle T_\alpha = 2 \times 5\,\mu\mathrm{s} = 10\,\mu\mathrm{s} \)