A nucleus A, with a finite de-broglie wavelength $$\lambda_A$$, undergoes spontaneous fission into two nuclei B and C of equal mass. 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:

We are told that the initial nucleus A has a finite de-Broglie wavelength $$\lambda_A$$. The de-Broglie relation states first of all

$$\lambda \;=\;\frac{h}{p},$$

where $$h$$ is Planck’s constant and $$p$$ is the linear momentum of the particle concerned.

Let the mass of each of the two daughter nuclei B and C be $$m$$. Because A splits into two equal-mass fragments, the mass of A is $$2m$$, but we shall not actually need that value; we only need equality of the masses of B and C.

Choose the positive $$x$$-direction as the direction in which A is moving just before fission. Denote the magnitudes of the velocities of B and C by $$v_B$$ and $$v_C$$ respectively. The statement “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” translates into

$$v_C \;=\;\frac{v_B}{2},\qquad$$ and the velocity of C is opposite in sign to that of B.

Thus the momenta of the three nuclei are

$$p_A \;=\;p_A\;(\text{along }+x),$$

$$p_B \;=\;m\,v_B\;(\text{also along }+x),$$

$$p_C \;=\;-\,m\,v_C \;=\;-\,m\left(\frac{v_B}{2}\right)\;(\text{along }-x).$$

Spontaneous fission occurs in free space, so linear momentum is conserved. Therefore

$$p_A \;=\;p_B + p_C.$$

Substituting the expressions just written, we have

$$p_A \;=\;m\,v_B \;-\;m\left(\frac{v_B}{2}\right).$$

Simplifying the right-hand side step by step,

$$p_A \;=\;m\,v_B \;-\;\frac{m\,v_B}{2} \;=\;\frac{m\,v_B}{2}.$$

Now compare this result with the momentum of B:

$$p_B \;=\;m\,v_B \;=\;2\,p_A.$$

Similarly, the magnitude of the momentum of C is

$$|p_C| \;=\;m\left(\frac{v_B}{2}\right) \;=\;p_A.$$

With all three momenta expressed in terms of $$p_A$$, we can write their de-Broglie wavelengths by applying $$\lambda = \dfrac{h}{p}$$ individually:

For A: $$\lambda_A \;=\;\frac{h}{p_A}.$$

For B: $$\lambda_B \;=\;\frac{h}{p_B} \;=\;\frac{h}{2\,p_A} \;=\;\frac{\lambda_A}{2}.$$

For C: $$\lambda_C \;=\;\frac{h}{|p_C|} \;=\;\frac{h}{p_A} \;=\;\lambda_A.$$

Hence the de-Broglie wavelengths of B and C are respectively

$$\lambda_B \;=\;\frac{\lambda_A}{2},\qquad \lambda_C \;=\;\lambda_A.$$

These correspond exactly to Option B (option number 2 in the list).

Hence, the correct answer is Option 2.