A 238U nucleus is moving in the x direction at 5.0 × 105 m/s when it decays into

Question

A 238U nucleus is moving in the x direction at 5.0 × 105 m/s when it decays into an alpha particle 4He and a 92 2

234Th. If the alpha particle moves off at 31.2° above the x axis with a speed of 1.4 × 107, what is the recoil velocity of the 90

thorium nucleus? Assume the uranium-thorium-alpha system is isolated; you may also assume the particles are pointlike.

Part (a) Write an expresssion for the conservation of momentum in the x direction. Your answer should be in terms of the masses m, mTh, the velocities v, vTh, and angles , Th (the angle of the thorium should be measured downward from the x-axis).

Expression :
muvu = __________________________________________

Select from the variables below to write your expression. Note that all variables may not be required. cos(), cos(), cos(), cos(), cos(Th), sin(), sin(), sin(), sin(), sin(Th), m, mTh, t, v, vTh

Part (b) Write an expression for the conservation of momentum in the y direction, taking upwards as positive. Expression :
0 = __________________________________________

Select from the variables below to write your expression. Note that all variables may not be required. cos(), cos(), cos(), cos(), cos(Th), sin(), sin(), sin(), sin(), sin(Th), m, mTh, t, v, vTh

Part (d) Find the numerical value of Th in degrees. You may assume that the mass of these particles is the number of nucleons (4 or 238) times the mass of a proton, 1.673 × 10-27 kg.

Numeric : A numeric value is expected and not an expression. Th = __________________________________________

Part (e) Write an expression for the velocity of the thorium nucuelus in terms of sin(Th). Expression :
vTh = __________________________________________

Select from the variables below to write your expression. Note that all variables may not be required. cos(), cos(), cos(), cos(), cos(Th), sin(), sin(), sin(), sin(), sin(Th), m, mTh, t, v, vTh

Explanation / Answer

Mass of Thorium nucleus, mTh = 234 U
Mass of Uranium nucleus, mU = 238 U
= 31.2 degrees
vU = 5 x 105 m/s
v = 140 x 105 m/s

a)
Momentum before decay in the x direction = mU x vU
= 238 x 5 x 105 = 11.9 x 107
Momentum after decay in the x direction = m V cos() + mTh x cos(Th)
Equating the above terms,
11.9 x 107 = m v cos() + mTh x vTh x cos(Th) ...(1)

b)
Momentum before decay in the y-direction = 0
Momentum after decay in the y-direction = m V sin() - mTh x vTh x sin(Th)
Since momentum is conserved in the y-direction,
0 = m V sin() - mTh x vTh x sin(Th) ...(2)

c)
From (1),
mTh x vTh x cos(Th) = 11.9 x 107 - m v cos()
234 x vTh cos(Th) = 11.9 x 107 - 4 x 1.4 x 107 x cos(31.2)
vTh cos(Th) = 3.04 x 105   ...(3)

From (2)
m V sin() = mTh x vTh x sin(Th)
4 x 1.4 x 107 x sin(31.2) = 234 x vTh sin(Th)
vTh sin(Th) = 1.24 x 105 ...(4)

(4)/(3)
tan(Th) = 1.24 / 3.04
Th = tan-1(0.40987)
= 22.19 degrees

e)
From (2)
m V sin() = mTh x vTh x sin(Th)
vTh = [m V sin()] / [mTh x sin(Th)]

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