Anonymous asked in Science & MathematicsPhysics · 11 months ago

A proton is launched from point 1 in with an initial velocity of 3.9×10^5 m/s.?

By how much has its kinetic energy changed, in eV, by the time it passes through point 2? Assume negative answer if kinetic energy decreases and positive if kinetic energy increases.


My Analysis :

KE(I) = 1/2 mv^2 ;

mass of proton = 1.672*10^-27kg

KE(I) = 3.26*10^-22J

At 100 V , delta(V) = W1/q ;

W1 = V*q = 100V*1.602*10^-19C=1.602*10^-17J

W2 = V*q (from 100 to 200) =


W1+W2 = 3.2*10^-17J


3.2*10^-17J - 1.602*10^-17J =

1.602*10^-17J or 199.975 eV

Is this correct, please help.

Attachment image

1 Answer

  • NCS
    Lv 7
    11 months ago
    Favourite answer

    Close. You've over-calculated.

    The amount by which it has CHANGED is -3.2e-17 J = -200 V.

    And you did not have to do two separate changes in potential.

    Across the change in potential (200V), the work done against the field is

    W = q*V = 1.6e-19C * 200V = 3.2e-17 J

    Besides, you should consider why we use the quantity "eV."

    The particle has a charge of -1e (where e = 1.6e-19 C)

    and so

    ΔKE = -1e * 200V = -200 V

    Easy peasy.

    Hope this helps!

Still have questions? Get answers by asking now.