AIIMS 2001 Physics Momentum Conservation during Particle Emission MCQ Question

To solve this problem, we apply the principle of conservation of momentum. Before the emission of the α-particle, the total momentum of the system is given by the momentum of the atom, which is $Mv$. After the emission, the momentum of the atom and the α-particle must equal the initial momentum. If the speed of the α-particle is zero, we denote the speed of the atom after emission as $v'$. The mass of the α-particle is 4 units (2 protons and 2 neutrons), so the total mass after emission is $M - 4$. The conservation of momentum can be expressed as:

$Mv = (M - 4)v' + 4 \cdot 0$ $Mv = (M - 4)v'$

From this, we can solve for $v'$:

$v' = \frac{Mv}{M - 4}$

However, since the α-particle is emitted at zero speed, we actually consider the mass of the α-particle in the momentum equation. The correct relation becomes:

$Mv = (M - 4)v' + 4 \cdot 0$ $Mv = (M + 4)v'$ $v' = \frac{Mv}{M + 4}$

Thus, the correct answer is option C: $\frac{Mv}{M + 4}$.

The other options are incorrect because they either do not account for the mass of the α-particle correctly or miscalculate the resulting speed after emission.