- #1
Mspike6
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Hey..
i got this question in a unit assessment test... i solved it (or atleast i think so :P ) but the answer doesn;t make sense . so i want someone to tell me if i did something wrong.
A plutonium-239, is initialy at rest, undergoes an alpha decay \, to produce a uranium 235 nucleus. the uranium-235 nucleus has a mass of 3.90*10-25 Kg, and moves away from the location of the decay at speed of 2.62 *105 m/s
Determine the minimum electric potential difference that is required to bring the alpha particle to rest.
Marks will be awarded based on the physics principles you provide, the formulas you state, the substitutions you show, and your final answer. (12 marks)
Solution:
To bring the alpha particle to rest, we must apply Electric Force equal to the alpha particle’s Kinetic energy.
qE= ½ mv2 –(1)
V=E/q
E= Vq –(2)
Substitute (2) into (1)
Vq2 = ½mv2
V= ½ mv2/q2
m=6.65*10-27Kg
q= 3.2*10-19 C
V= 2.3*1021 V
i got this question in a unit assessment test... i solved it (or atleast i think so :P ) but the answer doesn;t make sense . so i want someone to tell me if i did something wrong.
A plutonium-239, is initialy at rest, undergoes an alpha decay \, to produce a uranium 235 nucleus. the uranium-235 nucleus has a mass of 3.90*10-25 Kg, and moves away from the location of the decay at speed of 2.62 *105 m/s
Determine the minimum electric potential difference that is required to bring the alpha particle to rest.
Marks will be awarded based on the physics principles you provide, the formulas you state, the substitutions you show, and your final answer. (12 marks)
Solution:
To bring the alpha particle to rest, we must apply Electric Force equal to the alpha particle’s Kinetic energy.
qE= ½ mv2 –(1)
V=E/q
E= Vq –(2)
Substitute (2) into (1)
Vq2 = ½mv2
V= ½ mv2/q2
m=6.65*10-27Kg
q= 3.2*10-19 C
V= 2.3*1021 V