Question

A stationary thorium nucleus (A=220,Z=90) emits an alpha particle with kinetic energy ${{E}_{\alpha }}$. What is the kinetic energy of the recoiling nucleus?
$A.\quad \dfrac{{{E}_{\alpha }}}{108}$
$B.\quad \dfrac{{{E}_{\alpha }}}{110}$
$C.\quad \dfrac{{{E}_{\alpha }}}{55}$
$D.\quad \dfrac{{{E}_{\alpha }}}{54}$

Answer 1

Hint: To solve this problem, we will require the knowledge of conservation of momentum and conservation of kinetic energy. The conservation of momentum states that the initial and final momentum are equal, while the conservation of kinetic energy states, that the initial and final kinetic energies are equal.

Step by step solution:
Let’s start by making the interaction that occurs by thorium that disintegrates into an alpha particle and radium. The interaction is as follows, $_{90}^{220}Th\to _{2}^{4}He+_{88}^{216}Ra$. Here, the alpha particle is $_{2}^{4}He$.

Let’s solve the problem now. We know that the Thorium nucleus is at rest (stationary). Hence the initial velocity of LHS is zero. This implies, that the disintegrating nuclei, travels in opposite directions due to conservation of momentum.

Let the mass of thorium be $({{m}_{th}})$ , mass of alpha particle be $({{m}_{\alpha }})$ and the mass of radium be $({{m}_{ra}})$. We know that the initial velocity of thorium is zero. Let’s consider the initial velocity of the alpha particle be $({{v}_{\alpha }})$ and the initial velocity of radium be $({{v}_{ra}})$.

Using the conservation of momentum we get, initial momentum = final momentum.

That is, ${{m}_{th}}\times 0=({{m}_{\alpha }}\times {{v}_{\alpha }})+({{m}_{ra}}\times {{v}_{ra}})\Rightarrow {{m}_{\alpha }}{{v}_{\alpha }}=-{{m}_{ra}}{{v}_{ra}}\Rightarrow \dfrac{{{v}_{ra}}}{{{v}_{\alpha }}}=-\dfrac{{{m}_{\alpha }}}{{{m}_{ra}}}$.

The mass of the nuclei is given by the mass number A.

The negative sign only implies that the two nuclei travel in opposite directions. Hence, considering the modulus of the above equation we get, \[\Rightarrow \left| \dfrac{{{v}_{ra}}}{{{v}_{\alpha }}} \right|=-\dfrac{{{m}_{\alpha }}}{{{m}_{ra}}}\Rightarrow \dfrac{{{v}_{ra}}}{{{v}_{\alpha }}}=\left| -\dfrac{{{m}_{\alpha }}}{{{m}_{ra}}} \right|=\dfrac{4}{216}\Rightarrow \dfrac{{{v}_{ra}}}{{{v}_{\alpha }}}=\dfrac{1}{54}\].
Now, finding out the ratio of the kinetic energy of the radium is to the kinetic energy of alpha particle is given by, $\dfrac{{{E}_{radium}}}{{{E}_{alpha}}}=\dfrac{\dfrac{1}{2}{{m}_{ra}}v_{ra}^{2}}{\dfrac{1}{2}{{m}_{\alpha }}v_{\alpha }^{2}}=(\dfrac{{{m}_{ra}}}{{{m}_{\alpha }}}){{(\dfrac{{{v}_{ra}}}{{{v}_{\alpha }}})}^{2}}$. Here we will substitute in the ratio of the velocities. Hence, the ratio of the kinetic energies become, $\dfrac{{{E}_{radium}}}{{{E}_{alpha}}}=(\dfrac{{{m}_{ra}}}{{{m}_{\alpha }}}){{(\dfrac{{{v}_{ra}}}{{{v}_{\alpha }}})}^{2}}=(\dfrac{216}{4}){{(\dfrac{1}{54})}^{2}}\Rightarrow \dfrac{{{E}_{radium}}}{{{E}_{alpha}}}=54{{(\dfrac{1}{54})}^{2}}=\dfrac{1}{54}$.
Therefore, the kinetic energy of the recoiling nucleus (Radium) is, ${{E}_{radium}}=\dfrac{{{E}_{\alpha }}}{54}.$

Note:
An alternate way of doing the same problem is directly using the momentum conservation only. That is, ${{p}_{th}}={{p}_{\alpha }}+{{p}_{ra}}$. Since the thorium nucleus is at rest, hence its momentum is equal to zero. Therefore, momentum conservation equation becomes, $0={{p}_{\alpha }}+{{p}_{ra}}\Rightarrow {{p}_{\alpha }}=-{{p}_{ra}}=\left| p \right|$.

Let’s consider the magnitude of the equal and opposite momentum to be p. We know that the kinetic energy is given by, $E=\dfrac{{{p}^{2}}}{2m}$. Hence, the ratio of the kinetic energies of the disintegrated nuclei, radium and alpha particle is, \[\dfrac{{{E}_{radium}}}{{{E}_{\alpha }}}=\dfrac{\dfrac{{{p}^{2}}}{2{{m}_{ra}}}}{\dfrac{{{p}^{2}}}{2{{m}_{\alpha }}}}=\dfrac{{{m}_{\alpha }}}{{{m}_{ra}}}=\dfrac{4}{216}=\dfrac{1}{54}\].
Therefore, ${{E}_{radium}}=\dfrac{{{E}_{\alpha }}}{54}.$