Q. After absorbing a slowly moving neutron of mass $m_{N}$(momentum $\sim$0) a nucleus of mass $M$ breaks into two nuclei of masses $m_{1}$ and $5m_{1}$ ($6m_{1} = \ M\ + \ m_{N}$), respectively. If the de-Broglie wavelength of the nucleus with mass $m_{1}\ is\ \lambda$ , then de- Broglie wavelength of the other nucleus will be
A) 25$\lambda$
B) 5$\lambda$
C) $\frac{\lambda}{5}$
D) $\lambda$
Explanation
(d) de-Broglie wavelength $\ \lambda = \frac{- h}{mv} = \frac{h}{p}$
Where $p$ =momentum
By conservation of momentum
$\ \ \ \ \ \ \ \ \ \ \ \ \ P_{1} + P_{2}\ = 0$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P_{1}\ = \ P_{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\ \lambda}_{1} = \lambda_{2} = \ \lambda$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.