A nucleus of mass M at rest splits into two parts having masses \fracM^\prime3 and 2M' \over 3(M' \lt M) . The ratio of de Broglie wavelength of two parts will be :

Atoms and Nuclei PYQ — JEE Main 2022 · 26 Jul · Shift 2 · Q48 with Solution

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Correct answer: C

Given: A nucleus at rest splits into two fragments of masses $m_1=\frac{M'}{3}, \qquad m_2=\frac{2M'}{3}$ where $M'<M$ .
Use conservation of momentum: Since the original nucleus is at rest, its initial momentum is zero. Therefore after splitting, $\vec p_1+\vec p_2=0$ so the two fragments move in opposite directions with equal magnitudes of momentum: $p_1=p_2$
de Broglie wavelength formula: The de Broglie wavelength is $\lambda=\frac{h}{p}$

Since both fragments have the same momentum magnitude, $\lambda_1=\frac{h}{p_1}, \qquad \lambda_2=\frac{h}{p_2}$ and because $p_1=p_2$ , $\lambda_1=\lambda_2$
Ratio of wavelengths: $\lambda_1:\lambda_2 = 1:1$
Option check:
- A: $1:2$ ❌
- B: $2:1$ ❌
- C: $1:1$ ✅
- D: $2:3$ ❌

Therefore, the correct answer is C.