Question:

A nucleus A, with a finite de-broglie wavelength \(\lambda_A\), undergoes spontaneous fission into two nuclei B and C of equal mass. B flies in the same direction as that of A, while C flies in the opposite direction with a velocity equal to half of that of B. The de-Broglie wavelengths \(\lambda_B\) and \(\lambda_C\) of B and C are respectively :

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Apply conservation of momentum to find the momenta of B and C in terms of the momentum of A, then use \(\lambda = h/p\).
Updated On: Nov 5, 2026
  • \(\lambda_A,\ 2\lambda_A\)
  • \(2\lambda_A,\ \lambda_A\)
  • \(\lambda_A,\ \lambda_A/2\)
  • \(\lambda_A/2,\ \lambda_A\)
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The Correct Option is D

Solution and Explanation

Let A have mass \(2m\) and velocity \(V\), so its momentum is \(p = 2mV\) and \(\lambda_A = h/p\). B and C each have mass \(m\).
Let B move with velocity \(v\) along the direction of A; C then moves with velocity \(-v/2\).
Momentum conservation: \(2mV = mv - m\frac{v}{2} = \frac{mv}{2}\), so \(v = 4V\).
\(p_B = m\cdot 4V = 4mV = 2p\), so \(\lambda_B = \frac{h}{2p} = \frac{\lambda_A}{2}\).
\(p_C = m\left(-\frac{4V}{2}\right) = -2mV = -p\), so \(|p_C| = p\) and \(\lambda_C = \frac{h}{p} = \lambda_A\).
Answer: (4) \(\lambda_A/2,\ \lambda_A\)
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