Question

Points A and B have potentials of +250 V and-150 V respectively. An alpha particle (a helium nucleus that contains two proton
6) 8.ox 102 eV 7) (a)-1.1x 10*V (b)-38J
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Answer #1

Problem - 6

Using conservation of energy

Gain in kinetic energy = Loss in electric potential energy

Thus

KE = q(Delta V)

KE= (2 × 1.6 × 10-19 ) (250-(-150))

which gives us

KE 1.28 x 10-16J

and the energy in electron volts will be

16 1.28 x 10-1 1.6 x 10-19

which gives us

KE -800e

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