A proton and an α -particle enter a uniform magnetic field perpendicularly with the same speed. If proton takes 25 μ s to make 5 revolutions, then the periodic time for the α -particle would be

Question

A proton and an α -particle enter a uniform magnetic field perpendicularly with the same speed. If proton takes 25 μ s to make 5 revolutions, then the periodic time for the α -particle would be (A) 50 μ s (B) 25 μ s (C) 10 μ s (D) 5 μ s

Tardigrade Admin · Accepted Answer

Time period of proton Tp=(25/5)=5 μ s By using T=(2 π m/q B) ⇒ (Tα/Tp)=(mα/mp) × (qp/qα) =(4 mp/mp) × (qp/2 qp)=2 ⇒ Tα=2 Tp=10 μ s

Q. A proton and an $α$ -particle enter a uniform magnetic field perpendicularly with the same speed. If proton takes $25 μ s$ to make $5$ revolutions, then the periodic time for the $α$ -particle would be

$50 μ s$

$25 μ s$

$10 μ s$

$5 μ s$

Time period of proton $T_{p} = \frac{25}{5} = 5 μ s$
By using $T = \frac{2 π m}{qB}$
$\Rightarrow \frac{T _{α}}{T _{p}} = \frac{m _{α}}{m _{p}} \times \frac{q _{p}}{q _{α}}$
$= \frac{4 m _{p}}{m _{p}} \times \frac{q _{p}}{2 q _{p}} = 2$
$\Rightarrow T_{α} = 2 T_{p} = 10 μ s$

Q. A proton and an α -particle enter a uniform magnetic field perpendicularly with the same speed. If proton takes 25μs to make 5 revolutions, then the periodic time for the α -particle would be

50μs

25μs

10μs

5μs