Question

An alpha - particle and a proton are simultaneously accelerated from rest through a potential difference of 0.5 Mv, workdone by electric field is​

Answer 1

Given

Alpha - particle and a proton pass through a potential difference of 0.5 Mv

To Find

The work done by electric field

Solution

V = Voltage = 0.5 MV

$q_p$ = Charge of proton = $1.6\times 10^{-19}\ \text{C}$

$q_a$ = Charge of alpha particle = $2q_p=2\times1.6\times 10^{-19}\ \text{C}$

Work done by the electric field on proton

$E_p=q_pV\\\Rightarrow E_p=1.6\times 10^{-19}\times 0.5\times 10^6\\\Rightarrow E_p=8\times 10^{-14}\ \text{J}$

Work done by the electric field on the proton is $8\times 10^{-14}\ \text{J}$

Work done by the electric field on alpha particle

$E_a=q_aV\\\Rightarrow E_a=2\times 1.6\times 10^{-19}\times 0.5\times 10^6\\\Rightarrow E_a=1.6\times 10^{-13}\ \text{J}$

Work done by the electric field on the alpha particle is $1.6\times 10^{-13}\ \text{J}$

Answer 2

Explanation:

W=W1+W2

W1=q1V W2=q2V

W1 FOR ALPHA 2HE⁴

W2 FOR PROTON 1H¹

BY CALCULATING

0.5*1.6*10-13

1EV=1.6*10-¹⁹J

W=1+0 5= 1.5Mev

i skipped the calculation process hope you do it