Question

After absorbing a slowly moving neutron of mass $m_n$ (momentum =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 λ then de-Broglie wavelength of the other nucleus will be
(a) 25λ
(b) 5λ
(c) λ/5
(d) λ

Answer 1

Hint: Initially the neutron absorbed has zero momentum, since the mass of the neutron is approximately equal to the mass of the proton, so its mass cannot be zero and hence its velocity is zero. Also, the nucleus of mass M breaks into two nuclei and this happens due to the internal forces. We can use the concept of De Broglie wavelength relationship to arrive at a meaningful solution.

Complete step by step answer:
From the De Broglie wavelength relationship $\lambda ={\displaystyle \frac{h}{mv}}$where m is the mass of the moving body and h is Planck’s constant. Also, momentum of a body of mass, m moving with a velocity, v is given by $p=mv$.
$\Rightarrow \lambda ={\displaystyle \frac{h}{mv}}={\displaystyle \frac{h}{p}}$--(1)
Also, the breakup takes place due to internal forces and so momentum must remain conserved.
$\Rightarrow p_1+p_2=0$
$\Rightarrow p_1=p_2$
Using equation (1), we get,
$\Rightarrow {\displaystyle \frac{h}{p_1}}={\displaystyle \frac{h}{p_2}}$
$\therefore {\lambda }_1={\lambda }_2=\lambda$

So, the correct option is D.

Note:The wavelength is dependent upon the frequency and the speed of the propagating wave. Frequency is the characteristic of the source which is producing the wave. The SI unit of the wavelength is metre and velocity of the wave is m/s while for the frequency it is to be taken in Hertz (hz) always. Also, the law of conservation of momentum has its roots in law of conservation of energy.