An alpha particle (4He) undergoes an elastic collision with a stationary uranium nucleus (235U). What % of the kinetic E of the alpha particle is transferred to the uranium nucleus? collision is 1D

Answer 1

Let

m = 4 amu be the mass of the alpha particle,

M = 235 amu be the mass of the uranium atom,

v (unknown) be the initial velocity of the alpha particle,

w (unknown) be the final velocity of the alpha particle,

W (unknown) be the final velocity of the uranium atom.

The initial kinetic energy of the alpha particle is mv²/2.

The final kinetic energy of the uranium atom is MW²/2.

We are to compute their ratio:

(MW²/2) / (mv²/2) = (M/m)(W/v)² (0)

For that we will compute the fraction W/v.

Assuming no external forces, there is conservation of momentum:

the total momentum before the collision is equal the total momentum after:

mv = mw + MW (1)

Since the collision is elastic, energy is also conserved

mv²/2 = mw²/2 + MW²/2 (2)

From (1) express w:

w = v - MW/m

and substitute that into (2):

mv²/2 = m(v - MW/m)²/2 + MW²/2

Simplify that into

mv² = m(v - MW/m)² + MW²

mv² = mv² - 2MWv + M²W²/m + MW²

0 = -2MWv + M²W²/m + MW²

0 = -2v + MW/m + W

2 = (M/m)(W/v) + W/v

W/v = 2/(M/m + 1) = 2m/(M+m)

Substitute into (0)

(M/m)(W/v)² = (M/m)(2m/(M+m))² = 4Mm/(M+m)² =

4×235×4/239² = 0.065825 = 6.5825%