A nucleus disintegrates into two smaller parts, which have their velocities in the ratio 32. The ratio of their nuclear

Question

A nucleus disintegrates into two smaller parts, which have their velocities in the ratio 32. The ratio of their nuclear sizes will be ( x3)^1/3. The value of x is:

Accepted Answer

Given: A nucleus disintegrates into two parts with velocity ratio

\frac{v_1}{v_2} = \frac{3}{2}

Find: The value of $x$ if the ratio of nuclear sizes is $\left(\frac{x}{3}\right)^{1/3}$ .

Two daughter nuclei move in opposite directions after disintegration, labeled m1 with velocity v1 to the left and m2 with velocity v2 to the right, with v1 by v2 equal to 3 by 2.

Using conservation of momentum for the two fragments:

m_1 v_1 = m_2 v_2

Therefore,

\frac{m_2}{m_1} = \frac{v_1}{v_2} = \frac{3}{2}

Since nuclear mass density is constant, mass is proportional to volume. Hence,

\frac{m_1}{m_2} = \frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3}

So,

\left(\frac{r_2}{r_1}\right)^3 = \frac{m_2}{m_1}

Thus,

\left(\frac{r_2}{r_1}\right)^3 = \frac{3}{2}

Taking cube root,

\frac{r_2}{r_1} = \left(\frac{3}{2}\right)^{1/3} = \left(\frac{2}{3}\right)^{-1/3}

Matching with the given form $\left(\frac{x}{3}\right)^{1/3}$ , we get

x = 2

Therefore, the required value is $2$ .

Given: v1v2 = 32 Find: The value of x in the size ratio expression. From momentum conservation in two-body breakup, m1 v1 = m2 v2 so masses are inversely proportional to velocities. Hence, m2m1 = v1v2 = 32 For nuclei, radius varies as the cube root of mass: r m^1/3 Therefore, r2r1 = ( m2m1)^1/3 = ( 32)^1/3 Comparing with ( x3)^1/3 gives x = 2 The correct answer is 2.

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