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Ludmilka [50]
3 years ago
5

To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T

he proton must impact the nucleus with a kinetic energy of 1.80 MeV. Assume the nucleus remains at rest.
Required:
a. With what speed must the proton be fired toward the target?
b. Through what potential difference must the proton be accelerated from rest to acquire this speed?
Physics
1 answer:
vladimir1956 [14]3 years ago
3 0

Answer:

Explanation:

kinetic energy required = 1.80 MeV

= 1.8 x 10⁶ x 1.6 x 10⁻¹⁹ J

= 2.88 x 10⁻¹³ J

If v be the velocity of proton

1/2 x mass of proton x v² = 2.88 x 10⁻¹³

= .5 x 1.67 x 10⁻²⁷ x v² = 2.88 x 10⁻¹³

v² = 3.45 x 10¹⁴

v = 1.86 x 10⁷ m /s

If V be the potential difference required

V x e = kinetic energy . where e is charge on proton .

V x 1.6 x 10⁻¹⁹ = 2.88 x 10⁻¹³

V = 1.8 x 10⁶ volt .

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v = λf................. Equation 1

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Masja [62]

Answer:38.66 m

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No.

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