A proton, a deuteron and an alpha particle are accelerated through same potential difference and then they enter a normal uniform magnetic field. The ratio of their kinetic energies will be

Question

A proton, a deuteron and an alpha particle are accelerated through same potential difference and then they enter a normal uniform magnetic field. The ratio of their kinetic energies will be (A)  2:1:3  (B)  1:1:2  (C)  1:1:1  (D)  1:2:4

Tardigrade Admin · Accepted Answer

When a particle with charge q passes through a potential difference V, then kinetic energy acquired by particle is K=q V Given, q1: q2: q3=1: 1: 2 and V1: V2: V3=1: 1: 1 ∴ K1: K2: K3=1: 1: 2

Q. A proton, a deuteron and an alpha particle are accelerated through same potential difference and then they enter a normal uniform magnetic field. The ratio of their kinetic energies will be

$2 : 1 : 3$

$1 : 1 : 2$

$1 : 1 : 1$

$1 : 2 : 4$

When a particle with charge $q$ passes through a potential difference $V$ , then kinetic energy acquired by particle is $K = q V$
Given, $q_{1} : q_{2} : q_{3} = 1 : 1 : 2$
and $V_{1} : V_{2} : V_{3} = 1 : 1 : 1$
$∴ K_{1} : K_{2} : K_{3} = 1 : 1 : 2$

Q. A proton, a deuteron and an alpha particle are accelerated through same potential difference and then they enter a normal uniform magnetic field. The ratio of their kinetic energies will be

2:1:3

1:1:2

1:1:1

1:2:4

When a particle with charge q passes through a potential difference V, then kinetic energy acquired by particle is K=qV Given, q1​:q2​:q3​=1:1:2 and V1​:V2​:V3​=1:1:1 ∴K1​:K2​:K3​=1:1:2

$2 : 1 : 3$

$1 : 1 : 2$

$1 : 1 : 1$

$1 : 2 : 4$

When a particle with charge $q$ passes through a potential difference $V$ , then kinetic energy acquired by particle is $K = q V$
Given, $q_{1} : q_{2} : q_{3} = 1 : 1 : 2$
and $V_{1} : V_{2} : V_{3} = 1 : 1 : 1$
$∴ K_{1} : K_{2} : K_{3} = 1 : 1 : 2$