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Question: A proton and an\(\alpha -\)particle enter a uniform magnetic field perpendicularly with the same spe...

A proton and anα−\alpha -particle enter a uniform magnetic field perpendicularly with the same speed. If proton takes 25μsec25 \mu \mathrm { sec } to make 5 revolutions, then the periodic time for the α−\alpha -particle would be

A

50μsec50 \mu \mathrm { sec }

B
C
D

5μsec5 \mu \mathrm { sec }

Answer
Explanation

Solution

Time period of proton Tp=255=5μsecT _ { p } = \frac { 25 } { 5 } = 5 \mu \mathrm { sec }

By using T=2πmqBT = \frac { 2 \pi m } { q B } ⇒\Rightarrow TαTp=mαmp×qpqα\frac { T _ { \alpha } } { T _ { p } } = \frac { m _ { \alpha } } { m _ { p } } \times \frac { q _ { p } } { q _ { \alpha } } =4mpmp×qp2qp= \frac { 4 m _ { p } } { m _ { p } } \times \frac { q _ { p } } { 2 q _ { p } }

⇒\Rightarrow Tα=2Tp=10μsecT _ { \alpha } = 2 T _ { p } = 10 \mu \mathrm { sec }