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.

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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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