After absorbing a slowly moving neutron of mass \( m_N \) (momentum nearly zero), a nucleus of mass M breaks into two nuclei of masses \( m_1 \) and \( 5m_1 \) (\( 6m_1=M+m_N \)), so that the two nuclei fly apart with equal and opposite momenta. If the de Broglie wavelength of the nucleus of mass \( m_1 \) is \( \lambda \), then the de Broglie wavelength of the other nucleus will be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \lambda \)
The de Broglie wavelength \( \lambda=h/p \) depends only on the magnitude of the momentum. The mass matters only through the momentum. The two nuclei fly apart with equal and opposite momenta. So both have the same magnitude of momentum \( p \):
\( |m_1\vec{v}_1|=|5m_1\vec{v}_2| \)
\( \begin{aligned} \lambda_1 &= \frac{h}{p} \\ &= \lambda \end{aligned} \)
\( \begin{aligned} \lambda_2 &= \frac{h}{p} \\ &= \lambda \end{aligned} \)
NCERT (2026-27):
Wave Nature of Matter – “… De Broglie proposed that the wave length λ associated with a particle of momentum p is given as”
\( \displaystyle \lambda=\frac{h}{p} \)\( \displaystyle =\frac{h}{mv} \qquad (11.5) \)