Question

A proton, a deuteron and an $\alpha $ particle accelerated through the same potential difference enter a region of uniform magnetic field, moving at right angles to B. What is the ratio of their kinetic energies?
(A) $2:1:1$
(B) $2:2:1$
(C) $1:2:1$
(D) $1:1:2$

Answer 1

Hint: When a charge particle is enter in a region of uniform magnetic field of potential difference V then the kinetic energy is given by
$K.E. = qV$
Where
q $ = $ charge of particle
V $ = $ potential difference

Complete step by step answer:
We know that the charge of proton ${q_p} = e$
Charge at deuteron ${q_d} = e$
Charge at $\alpha $ particle ${q_\alpha } = 2e$
Given that potential difference at which particles was accelerated is same i.e., V $(let)$
We also know that in this case the expression for kinetic energy of charged particles is
$K.E. = qV$
Hence
K.E. of proton $ = {q_p}V = eV$ …..(1)
K.E. of deuteron $ = {q_d}V = eV$ …..(2)
K.E. of $\alpha $ particle $ = {q_\alpha }V = 2eV$ …..(3)
So, the ratio of their K.E. is
$K{E_p} = K{E_d}:K{E_\alpha } = eV:eV:2eV$
$K{E_p}:K{E_d}:K{E_\alpha } = 1:1:2$

So, the correct answer is “Option D”.

Note:
If potential difference is same for all particles and magnetic field is perpendicular and magnetic field is perpendicular then only electric force is effecting in this condition and kinetic energy is directly proportional to charge of particle. i.e., $K.E.\propto q$