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Q.

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

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a

1 MeV

b

4 MeV

c

2 MeV

d

0 .5 MeV

answer is A.

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Detailed Solution

we know that $r = \frac{mv}{qB} or r^{2} = \frac{m^{2} v^{2}}{q^{2} B^{2}}$
In terms of energy, the above equation can be written as
$r^{2} = \frac{2 m \times (1 / 2 {mv}^{2})}{q^{2} B^{2}} = \frac{2 mE}{q^{2} B^{2}}$
Here r and B are same, hence
$\begin{array}{ll} \frac{m_{1} E_{1}}{{q_{1}}^{2}} = \frac{m_{2} E_{2}}{{q_{2}}^{2}} or \frac{m \times 1}{e^{2}} = \frac{3 m \times E_{2}}{4 e^{2}} \\ ∴ E_{2} = 1 MeV \end{array}$

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