Question

A particle of mass 4M at rest disintegrates into two particles of mass M and 3M respectively having non zero velocities. The ratio of de-Broglie wavelength of particle of mass M to that of mass 3M will be :

• A. 1 : 1
• B. 1 : 3
• C. 3 : 1
• D. 1 : $\sqrt{3}$

Hint:

In any internal explosion or disintegration from rest, the fragments always have equal and opposite momenta.
Consequently, their de-Broglie wavelengths are always identical.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
When a particle at rest disintegrates into two fragments, linear momentum must be conserved.
According to the Law of Conservation of Momentum:
Initial Momentum = Final Momentum
Step 2: Key Formula or Approach:
\[ \vec{P}_i = \vec{P}_1 + \vec{P}_2 \]
Since initial momentum \(\vec{P}_i = 0\):
\[ 0 = \vec{P}_1 + \vec{P}_2 \implies |\vec{P}_1| = |\vec{P}_2| \]
The magnitude of momentum of both fragments is equal.
Step 3: Detailed Explanation:
The de-Broglie wavelength is given by \(\lambda = \frac{h}{p}\).
For the fragment of mass \(M\), the wavelength is \(\lambda_1 = \frac{h}{P_1}\).
For the fragment of mass \(3M\), the wavelength is \(\lambda_2 = \frac{h}{P_2}\).
Since we established from conservation of momentum that \(P_1 = P_2\):
\[ \lambda_1 = \lambda_2 \]
The ratio \(\lambda_1 : \lambda_2 = 1 : 1\).
Step 4: Final Answer:
The ratio of the de-Broglie wavelengths is 1 : 1.