Question

During the disintegration of a radioactive nucleus of mass number 208 at rest, two alpha particles each with kinetic energy E are emitted. The total kinetic energy of the emitted alpha particles and the daughter nucleus after the disintegration is

• A. $\frac{51E}{25}$
• B. $\frac{51E}{50}$
• C. $\frac{52E}{25}$
• D. $\frac{26E}{25}$

Hint:

In a two-body decay ($P \to D + \alpha$), momentum is conserved: $\vec{p}_D = -\vec{p}_\alpha$. In a multi-body decay like this one, the daughter nucleus recoils to balance the vector sum of the momenta of the emitted particles. The maximum possible total kinetic energy occurs when the momenta of the emitted particles are aligned.

Answer 1

The Correct Option is C

Step 1: State the conservation laws.
The parent nucleus is at rest, so the total initial momentum is zero. Due to conservation of momentum, the total final momentum must also be zero. Let $M_p = 208$ be the mass number of the parent nucleus. The mass of a nucleus is proportional to its mass number.

Step 2: Define the masses and velocities of the final products.
The products are two alpha particles ($\alpha$) and a daughter nucleus ($D$). Mass number of an alpha particle: $M_\alpha = 4$. Mass number of the daughter nucleus: $M_D = 208 - 2(4) = 200$. Let $\vec{v}_{\alpha 1}, \vec{v}_{\alpha 2}, \vec{v}_D$ be the velocities, and $E_{\alpha 1} = E_{\alpha 2} = E$ be the kinetic energy of each alpha particle.

Step 3: Apply the conservation of momentum.
Since the parent nucleus is at rest, the initial momentum is zero. \[ \vec{p}_{initial} = 0 = \vec{p}_{final} = M_\alpha \vec{v}_{\alpha 1} + M_\alpha \vec{v}_{\alpha 2} + M_D \vec{v}_D. \] The two $\alpha$ particles are emitted, and their kinetic energies are equal. Since $KE = p^2/(2M)$, and their masses are equal, their momenta must be equal in magnitude: $p_{\alpha 1} = p_{\alpha 2}$. For momentum to be conserved, the two $\alpha$ particles must be emitted in opposite directions, and the daughter nucleus must recoil to balance the net momentum of the two $\alpha$'s.

Step 4: Analyze the momentum of the two alpha particles.
Since $p_{\alpha 1} = p_{\alpha 2}$, and the total momentum is zero, the net momentum of the two $\alpha$'s can be in any direction. The lowest kinetic energy for the daughter nucleus happens when the two $\alpha$'s are emitted in opposite directions. The highest kinetic energy for the daughter nucleus happens when the two $\alpha$'s are emitted in the same direction, which violates conservation of momentum. The total momentum of the two alpha particles is $\vec{P}_\alpha = \vec{p}_{\alpha 1} + \vec{p}_{\alpha 2}$. By conservation of momentum, $\vec{p}_D = -\vec{P}_\alpha$.
The magnitude of the momentum of an alpha particle is $p_\alpha = \sqrt{2 M_\alpha E} = \sqrt{2(4)E} = \sqrt{8E}$.
The possible range of $\vec{P}_\alpha$ is $[0, 2\sqrt{8E}]$. For maximum recoil, $p_{\alpha 1}$ and $p_{\alpha 2}$ must be parallel. \[ P_{\alpha, max} = p_{\alpha 1} + p_{\alpha 2} = 2\sqrt{8E}. \] The momentum of the daughter nucleus is $p_D = P_{\alpha}$.
The kinetic energy of the daughter nucleus is $E_D = \frac{p_D^2}{2 M_D}$.
For the maximum recoil, $E_{D, max} = \frac{(2\sqrt{8E})^2}{2(200)} = \frac{4(8E)}{400} = \frac{32E}{400} = \frac{2E}{25}$.

Step 5: Calculate the total kinetic energy.
The total kinetic energy is the sum of the kinetic energies of the two alpha particles and the daughter nucleus. \[ KE_{total} = E_{\alpha 1} + E_{\alpha 2} + E_D = E + E + E_D = 2E + E_D. \] Using the value $E_D = \frac{2E}{25}$ (which corresponds to the maximum possible recoil). \[ KE_{total} = 2E + \frac{2E}{25} = E \left(2 + \frac{2}{25}\right) = E \left(\frac{50+2}{25}\right) = \frac{52E}{25}. \] This corresponds to option (C).