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A uranium nucleus 238U may stay in one piece for billions of years, but sooner or later it decays into an α particle of mass 6.64×10−27 kg and 234Th nucleus of mass 3.88 × 10−25 kg, and the decay process itself is extremely fast (it takes about 10−20 s). Suppose the uranium nucleus was at rest just before the decay. If the α particle is emitted at a speed of 8.57×106 m/s, what would be the recoil speed of the thorium nucleus? Answer in units of m/s.

Respuesta :

Answer:

1.47*10^5m/s

Explanation:

Using law of conservation of momentum

Since the uranium was initially at rest then the total momentum (did not change) is conserved and its initial value is zero ; 0= M1*V1+M2V2

-M1V1 = M2V2 where V1 is the velocity of the alpha particle and m1 is its mass and M2 is the mass of the thorium and V2 is its velocity.

- 6.64*10^-27*8.56*10^6 = 3.88*V2

V2 = -5.69*10^-20/3.88*10^-25 = -1.47*10^5m/s the negative sign mean the vector velocity of the nucleus point in opposite direction to the particle.

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