Atoms & Nuclei

JEE Main level, single correct, medium

230 of 239

Question 230 of 239

An unstable heavy nucleus at rest breaks into two nuclei which move away with velocities in the ratio of 8:278: 27. The ratio of the radii of the nuclei (assumed to be spherical ) is:

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Answer: (C) 3:23: 2

Let heavy nucleus breaks into two nuclei of mass m1m_1 and m2m_2 and move away with velocities V1V_1 and V2V_2 respectively.
According to question, V1V2=827\frac{V_1}{V_2}=\frac{8}{27} m1V1=m2V2m_1 V_1=m_2 V_2 (Law of momentum conservation)
⇒m1m2=V2V1=278ρ×43πR13ρ×43πR23(∵ density ρ= mass  volume )\begin{aligned} &\Rightarrow \frac{m_1}{m_2}=\frac{V_2}{V_1}=\frac{27}{8} \\ &\frac{\rho \times \frac{4}{3} \pi R_1^3}{\rho \times \frac{4}{3} \pi R_2^3} \quad\left(\because \text { density } \rho=\frac{\text { mass }}{\text { volume }}\right) \end{aligned}

⇒(R1R2)=(278)13=(32)3×13∴R1R2=32\begin{aligned} &\Rightarrow\left(\frac{\mathrm{R}_1}{\mathrm{R}_2}\right)=\left(\frac{27}{8}\right)^{\frac{1}{3}}=\left(\frac{3}{2}\right)^{3 \times \frac{1}{3}} \\ &\therefore \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{3}{2} \end{aligned}

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