Question

A neutron moving with a speed of 10^6m/s suffers a head on collision with a nucleus of mass number 80.what is the fraction of energy retained by the nucleus

Answer 1

Since Neutron collides with the nucleus, momentum before the collision = u(m₁ - m₂) where u is the initial velocity of the protons = 10⁶ m/s.

∴ Momentum = 10⁶(m₁ - m₂)

After collision, let the speed of the Neutron after the collision be v.

∴ Momentum after collision = v(m₁ + m₂)

Now, Momentum will remains conserved during the collision,

∴ u(m₁ - m₂) = v(m₁ + m₂)

⇒ 10⁶(1 - 80) = v(1 + 80)   [Mass of the Proton = 1 u].

∴ v = 10⁶(1 - 80)/(1 + 80)

⇒ v/u = (1 - 80)/(1 + 80)   -------eq(i).

Now, Fraction of the Kinetic energy retained =  0.5mv²/0.5mu²

⇒ 0.5mv²/0.5mu² = (v/u)²

⇒ v²/u² = [(1 - 80)/(1 + 80)]²

⇒ (v/u)² = (-79)²/(81)²

⇒ (v/u)² = 6241/6561

∴ Fraction of the energy retained by the Nucleus is 6241/6561.

Hope it helps.

Answer 2

Answer:

Explanation:year it's correct