Question

A nucleus $A$, with a finite de Broglie wavelength ${\lambda }_A$, undergoes spontaneous fission into two nuclei $B$ and $C$ of equal masses. $B$ 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,

• A. ${\lambda }_A,\frac{{\lambda }_A}{2}$
• B. $\frac{{\lambda }_A}{2},{\lambda }_A$
• C. $2{\lambda }_A,{\lambda }_A$
• D. ${\lambda }_A,2{\lambda }_A$

Answer 1

The correct option is B $\frac{{\lambda }_A}{2},{\lambda }_A$

Let $P$ be the momentum of nucleus $\text{A}$. Therefore,

$P_A=P$ and $P_C=P_B/2$

Since only internal forces are involved, so we can apply conservation of linear momentum,

$\overrightarrow{P_A}=\overrightarrow{P_B}+\overrightarrow{P_C}$

$P=P_B-P_B/2=P_B/2$

$\therefore P_A=P;P_B=2P;P_C=P$

As, $\lambda =\frac{h}{P}\Rightarrow {\lambda }_A=\frac{h}{P_A}=\frac{h}{P}$

$\therefore {\lambda }_B=\frac{{\lambda }_A}{2}$ and ${\lambda }_C={\lambda }_A$

Hence, $\left(D\right)$ is the correct answer.