If an α-particle and a proton are accelerated from rest by a potential difference of 1 megavolt then the ratio of their kinetic energy will be

Question

If an α-particle and a proton are accelerated from rest by a potential difference of 1 megavolt then the ratio of their kinetic energy will be (A) (1/2) (B) 1 (C) 2 (D) 4

Tardigrade Admin · Accepted Answer

Δ KE =q V (Δ KE α/Δ KE p)=(qα V/qp V)=(qα/qp)=2

Q. If an $α$ -particle and a proton are accelerated from rest by a potential difference of $1$ megavolt then the ratio of their kinetic energy will be

$\frac{1}{2}$

1

2

4

$Δ K E = q V$
$\frac{Δ K E _{α}}{Δ K E _{p}} = \frac{q _{α} V}{q _{p} V} = \frac{q _{α}}{q _{p}} = 2$

Q. If an α-particle and a proton are accelerated from rest by a potential difference of 1 megavolt then the ratio of their kinetic energy will be

21​

1

2

4

ΔKE=qV ΔKEp​ΔKEα​​=qp​Vqα​V​=qp​qα​​=2

Q. If an $α$ -particle and a proton are accelerated from rest by a potential difference of $1$ megavolt then the ratio of their kinetic energy will be

$\frac{1}{2}$

$Δ K E = q V$
$\frac{Δ K E _{α}}{Δ K E _{p}} = \frac{q _{α} V}{q _{p} V} = \frac{q _{α}}{q _{p}} = 2$