Question
The nucleus of a certian atom 3.8X10^-25 Kg and is at rest. The nucleus is redioactive...
The nucleus of a certian atom 3.8X10^-25 Kg and is at rest.
The nucleus is redioactive and suddenly ejects from itself a particle of mass 6.6X10^-27 Kg and speed 1.5X10^7.
Apply the conservation of momentum before = Momentum after.
Find the recoil speed of the nucleus that is left behind.
Homework Answers
Answer #1
Mass of the nucleus (MN)=3.8*10-25Kg.
Initial Momentum = 0 (Nucleus is at rest Velocity is zero)
Momentum of the ejected particle = Mass of ejected particle * velocity
= 6.6*10-27*1.5*107=9.9*1020Kg-m/s
Final momentum = Momentum of ejected particle + Momentum of Nucleus.
According to conservation of momentum Initial momentum = final momentum
Final momentum = 0
Momentum
of ejected particle = -( Momentum of nucleus)
Momentum of nucleus = -9.9*10-20 Kg-m/s
Recoil
velocity of nucleus= -9.9*10-20/3.8*10-25=
-2.605*105m/s (-ve sign indicate nuclues will move in
opposite side with respect to the ejected particle)
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