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13.Nuclei
medium
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 $3m_1$ $(4m_1 = M + m_N)$, 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
$9 \lambda$
B
$3 \lambda$
C
$\frac{\lambda }{3}$
D
$\lambda$
Solution
since initial momentum of the system is $P_{i}=0$
Hence final momentum will also be $P_{f}=0$ from conservation of momentum
Now, let momentum of two nuclei produced be $P_{1}$ and $P_{2}$.
Then, both momentum will be opposite to each other and $P_{1}=P_{2}$ in magnitude
Now, $\lambda=\frac{h}{m v}=\frac{h}{p}$
since, $P_{1}=P-2 \Longrightarrow \lambda_{1}=\lambda_{2}=\lambda$
Standard 12
Physics
