Question

A particle of mass \( M \) at rest decays into masses \( m_1 \) and \( m_2 \) with non-zero velocities. The ratio of de Broglie wavelengths \( \lambda_1 \) and \( \lambda_2 \) of the particles is

• A. \( \frac{m_1}{m_2} \)
• B. \( \sqrt{\frac{m_1}{m_2}} \)
• C. \( \frac{m_2}{m_1} \)
• D. \( 1 : 1 \)

Hint:

If a particle at rest splits into two parts, their momenta are equal and opposite, so de Broglie wavelengths become equal.

Answer 1

The Correct Option is D

Step 1: Apply conservation of momentum.
Initial particle is at rest, so total initial momentum is zero.
\[ p_1 + p_2 = 0 \]

Step 2: Relation between momenta.
\[ p_1 = -p_2 \] Magnitude wise:
\[ |p_1| = |p_2| \]

Step 3: Use de Broglie wavelength formula.
\[ \lambda = \frac{h}{p} \]

Step 4: Write wavelength expressions.
\[ \lambda_1 = \frac{h}{p_1}, \quad \lambda_2 = \frac{h}{p_2} \]

Step 5: Taking ratio.
\[ \frac{\lambda_1}{\lambda_2} = \frac{p_2}{p_1} \]

Step 6: Substitute momentum equality.
Since \( |p_1| = |p_2| \), we get:
\[ \frac{\lambda_1}{\lambda_2} = 1 \]

Step 7: Final conclusion.
\[ \lambda_1 : \lambda_2 = 1 : 1 \] \[ \boxed{1 : 1} \] Hence, correct answer is option (D).