Anonymous
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.

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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) =

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

W1+W2 = 3.2*10^-17J

W=ΔKE

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

1.602*10^-17J or 199.975 eV

Is this correct, please help.

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1 Answer

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  • 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!

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