Question

Uranium -238, initially at rest, emits an alpha particle with a speed of $1.4 \times {10^7}m/s$. Calculate the recoil speed of the residual nucleus thorium -234. Assume that the mass of a nucleus is proportional to the mass number.

Answer 1

Hint: The momentum of the object is proportional to the product of mass (m) of the object in kg and velocity (v) of the moving object in m/s. The moving body follows the rule of conservation of momentum, where the initial momentum equals final momentum.

Formula used:
Momentum $(P) = mv $,
Where,
$m$ = mass of the object in $kg$
$v$ = velocity of object in $m/s $
Conservation of momentum, initial momentum = final momentum.

Complete step by step solution:
Let us first write the information given in the question.
Initial i.e. when means, u=0, speed of alpha particle=$1.4 \times {10^7}m/s$, we have to find the speed of the residual nucleus.
The uranium-238 undergoes α-decay to produce Thorium-234. The reaction of α-decay as shown below,$_{{\text{92}}}^{{\text{238}}}{\text{U}} \to _{{\text{90}}}^{{\text{234}}}{\text{Th + }}_{\text{2}}^{\text{4}}{\text{He}}$
During α-decay, the atomic nucleus undergoes the emission of alpha particle which is $_{\text{2}}^{\text{4}}{\text{He}}$, such that there is a loss of atomic number by 2 and atomic mass by 4 as shown.
Now, let us use the concept of conservation of momentum which says that the initial momentum of a system equals the final moment of the same system.
As the uranium nucleus is at rest initially hence, its initial momentum is zero i.e. Pinitial=0, final momentum Pfinal will be equal to the sum of an emitted alpha particle and residual nucleus (thorium).
This means if we apply the law of conservation of momentum we can write the following.
${P_{alpha - particle}} + {P_{thorium}} = 0$ (1)
We can rewrite the above equation as follows.
${P_{alpha - particle}} = - {P_{thorium}}$
Now, let us use the formula of momentum.
${(mv)_{alpha - particle}} = - {(mv)_{thorium}}$
Let us now write the expression for the velocity of the residual nucleus using the above expression.${v_{thorium}} = - \dfrac{{{{(mv)}_{alpha - particle}}}}{{{m_{thorium}}}}$
Let us now substitute the values, mass of thorium = 234, mass of alpha particle = 4 and velocity of alpha particle = 1.4 × 107m/s.
${v_{thorium}} = - \dfrac{{4 \times 1.4 \times {{10}^7}}}{{234}}$
Let us further simplify it.
${v_{thorium}} = - 2.393 \times {10^5}m/s$
Hence, the required recoil speed of thorium is $ - 2.393 \times {10^5}m/s$here negative sign indicates the direction.

Additional information:
Momentum is a vector quantity. It has a magnitude as well as direction.
The SI unit of momentum is kilogram meters per second (kg m/s).
Momentum for the many-particle systems can be written as,
$p = \sum\limits_{i = 1}^n {{m_i}} {v_i}$

Note: Recoil is a situation that occurs when initially object A is at rest and releases another object B of smaller mass which has an impact on this body A. in this case law of conservation of momentum needs to be satisfied. Hence, recoil speed is the speed of object A after the emission of object B.