A nucleus disintegrates into two nuclear parts, in such a way that ratio of their nuclear sizes is $1: 2^{1 / 3}$. Their respective speed have a ratio of $n: 1$. The value of $n$ is __________.
Answer (integer)
2
Solution
Let the masses of the two nuclear parts be $m_1$ and $m_2$, and their respective speeds be $v_1$ and $v_2$. According to the problem, the ratio of their nuclear sizes is $1 : 2^{1/3}$. Since the nuclear size is proportional to the cube root of the mass, we can write:
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$\frac{m_1}{m_2} = \left(\frac{1}{2^{1/3}}\right)^3 = \frac{1}{2}$
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Now, according to the conservation of linear momentum, the momentum before disintegration is equal to the momentum after disintegration:
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$m_1 v_1 = m_2 v_2$
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From the problem statement, the ratio of their respective speeds is $n : 1$, so we can write:
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$v_1 = n \cdot v_2$
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Substitute the expression for $v_1$ into the momentum conservation equation:
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$m_1 (n \cdot v_2) = m_2 v_2$
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We know the mass ratio, so substitute that into the equation:
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$\frac{1}{2} m_2 (n \cdot v_2) = m_2 v_2$
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Divide both sides by $m_2 v_2$:
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$\frac{1}{2} n = 1$
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Now, solve for $n$:
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$n = 2$
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Thus, the value of $n$ is 2.
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Nuclear Fission and Fusion
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