Part A How much voltage must be used to accelerate a proton (radius 1.2 ×10−15m) so that it has sufficient energy to just penetrate a silicon nucleus? A silicon nucleus has a charge of +14e, and its radius is about 3.6 ×10−15m. Assume the potential is that for point charges.

Step-by-step explanation:

Qp = charge on proton = 1.6 x 10-19 C

Qs = charge on silicon = 14 x 1.6 x 10-19 C

rf = final distance from nucleus = ∞

ri = initial distance from nucleus = (3.6 x 10-15 + 1.2 x 10-15 ) = 4.8 x 10-15 m

initial Potential energy is given as

Ui = k Qp Qs / ri = (9 x 109) (1.6 x 10-19 ) (14 x 1.6 x 10-19 ) / (4.8 x 10-15 ) = 6.72 x 10-13 J

final Potential energy is given as

Uf = k Qp Qs / rf = (9 x 109) (1.6 x 10-19 ) (14 x 1.6 x 10-19 ) / (∞) = 0 J

Change in Potential energy = ΔU = Ui - Uf = 6.72 x 10-13 - 0 = 6.72 x 10-13 J

Let the Voltage through which proton is accelerated = V

Energy gained due to potential difference = Qp V

Using conservation of energy

Qp V = 6.72 x 10-13

(1.6 x 10-19 ) V = 6.72 x 10-13

V = 4.2 x 106 volts

a

step-by-step explanation:

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