Question

Uranium 238 nucleus, initially at rest, emits an alpha particle with the speed of $1.4 \times {10^7}$ meter per second. Calculate the recoil speed of the residual nucleus Thorium 234 resumes that mass of the nucleus is proportional to the mass number.

Answer 1

Hint: The speed of residual nucleus can be found out by considering the initial velocity of the alpha partial as zero. Law of conservation of momentum is used to find the Final solution. That is, the initial momentum of the system is equal to the final momentum of the system.

Formula used:
The expression for finding the momentum of the object is Momentum $P=mv$, where $m$ is mass and $v$ is velocity.

Complete step by step answer:
Let us first write the information given in the question.
Initially at rest means, $u=0$, speed of alpha particle=$1.4 \times {10^7}m/s$, we have to find the speed of the residual nucleus.
Now, let us use the concept of conservation of momentum which says that initial momentum of a system is equal to the final moment of the same system.
As the uranium nucleus is at rest initially hence, its initial momentum is zero. And final momentum will be the sum of emitted alpha particles 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 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 × 10$^7$m/s.
${v_{thorium}} = - \dfrac{{4 \times 1.4 \times {{10}^7}}}{{234}}$
Let us further simplify it.
${v_{thorium}} = - 2.393 \times {10^5}m/s$

$\therefore $The required recoil speed of thorium is $ - 2.4 \times {10^5}m/s$, here negative sign indicates the direction.

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 the law of conservation of momentum needs to be satisfied. Hence, recoil speed is the speed of object A after collision of object B.
- Final momentum is taken as the sum of the velocity of the alpha particle and residual nucleus.