JEE MAIN - Physics (2021 - 18th March Evening Shift - No. 15)
A proton and an $$\alpha$$-particle, having kinetic energies Kp and K$$\alpha$$ respectively, enter into a magnetic field at right angles.
The ratio of the radii of trajectory of proton to that of $$\alpha$$-particle is 2 : 1. The ratio of Kp : K$$\alpha$$ is :
The ratio of the radii of trajectory of proton to that of $$\alpha$$-particle is 2 : 1. The ratio of Kp : K$$\alpha$$ is :
8 : 1
4 : 1
1 : 8
1 : 4
Explanation
Radius, $$r = {{mv} \over {qB}} = {{\sqrt {2mK} } \over {qB}}$$
$$ \Rightarrow K = {{{r^2}{q^2}{B^2}} \over {2m}}$$
$$ \therefore $$ $${{{K_p}} \over {{K_\alpha }}} = {\left( {{{{r_p}} \over {{r_\alpha }}}} \right)^2} \times {\left( {{{{q_p}} \over {{q_\alpha }}}} \right)^2} \times {{{m_\alpha }} \over {{m_p}}}$$
$$ = {\left( {{2 \over 1}} \right)^2} \times {\left( {{1 \over 2}} \right)^2} \times 4$$
$$ = 4$$
$$ \Rightarrow K = {{{r^2}{q^2}{B^2}} \over {2m}}$$
$$ \therefore $$ $${{{K_p}} \over {{K_\alpha }}} = {\left( {{{{r_p}} \over {{r_\alpha }}}} \right)^2} \times {\left( {{{{q_p}} \over {{q_\alpha }}}} \right)^2} \times {{{m_\alpha }} \over {{m_p}}}$$
$$ = {\left( {{2 \over 1}} \right)^2} \times {\left( {{1 \over 2}} \right)^2} \times 4$$
$$ = 4$$
Comments (0)
