Question

A proton of mass m and charge $+ e$ is moving in a circular orbit of a magnetic field with energy 1MeV. What should be the energy of α-particle (mass = 4 m and charge = +2e), so that it can revolve in the path of same radius

• A. 1 MeV
• B. 4 MeV
• C. 2 MeV
• D. 0.5 MeV

Answer 1

Answer: (A)

1 MeV

Answer 2

By using $r = \frac { \sqrt { 2 m K } } { q B }$ ; $r \rightarrow$ same, $B \rightarrow$same

$\Rightarrow$ $K \propto \frac { q ^ { 2 } } { m }$

Hence $\frac { K _ { \alpha } } { K _ { p } } = \left( \frac { q _ { \alpha } } { q _ { p } } \right) ^ { 2 } \times \frac { m _ { p } } { m _ { \alpha } }$ $= \left( \frac { 2 q _ { p } } { q _ { p } } \right) ^ { 2 } \times \frac { m _ { p } } { 4 m _ { p } } 1$

$\Rightarrow$ $K _ { \alpha } = K _ { p } = 1 \mathrm { meV }$