Question

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 $\lambda$, the de-Broglie wavelength of the other nucleus will be:

• A. $\lambda$
• B. $25\lambda$
• C. $5\lambda$
• D. $\lambda /5$

Answer 1

Answer: (A)

$\lambda$

Momentum should be conserved.

As initial momentum was $zero$ so final should also be zero.

so $p_1+p_2=0$ or $p_1=-p_2$ i.e. $both$ have $same$ values..............(1)

where $p_1$ is momentum of $m_1$ and $p_2$ is that of $m_2$

The deBroglie wavelength is given by $\lambda =\frac{h}{p}$

As both mass have same $momentum,$ , $p_1=-p_2$

so same values of $wavelength.$ So answer is $\lambda$