If an alpha particle and a proton are accelerated from rest by a potential difference of 1MeV , then the ratio of their kinetic energies will be

Question

If an alpha particle and a proton are accelerated from rest by a potential difference of 1MeV , then the ratio of their kinetic energies will be (A) 3 (B) 1 (C) 2 (D) 4

Tardigrade Admin · Accepted Answer

From conservation of mechanical energy Δ KE=q(Vi - Vf) (Kα /Kp)=(2 e (V)/e (V)) (Kα /Kp)=(2/1)

Q. If an alpha particle and a proton are accelerated from rest by a potential difference of $1 M e V$ , then the ratio of their kinetic energies will be

$3$

$1$

$2$

$4$

From conservation of mechanical energy
$Δ K E = q (V_{i} - V_{f})$
$\frac{K _{α}}{K _{p}} = \frac{2 e ( V )}{e ( V )}$
$\frac{K _{α}}{K _{p}} = \frac{2}{1}$

Q. If an alpha particle and a proton are accelerated from rest by a potential difference of 1MeV , then the ratio of their kinetic energies will be

3

1

2

4

From conservation of mechanical energy ΔKE=q(Vi​−Vf​) Kp​Kα​​=e(V)2e(V)​ Kp​Kα​​=12​

Q. If an alpha particle and a proton are accelerated from rest by a potential difference of $1 M e V$ , then the ratio of their kinetic energies will be

$3$

$1$

$2$

$4$

From conservation of mechanical energy
$Δ K E = q (V_{i} - V_{f})$
$\frac{K _{α}}{K _{p}} = \frac{2 e ( V )}{e ( V )}$
$\frac{K _{α}}{K _{p}} = \frac{2}{1}$