An unstable heavy nucleus at rest breaks into | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the heavy nucleus break into two nuclei of masses $m_1$ and $m_2$ moving with velocities $V_1$ and $V_2$ respectively.According to the law of

Vedclass · Accepted Answer

$3 : 2$

Vedclass · Answer

(C) Let the heavy nucleus break into two nuclei of masses $m_1$ and $m_2$ moving with velocities $V_1$ and $V_2$ respectively.According to the law of conservation of linear momentum,the initial momentum is zero,so $m_1 V_1 = m_2 V_2$.Given the ratio of velocities $\frac{V_1}{V_2} = \frac{8}{27}$,we have $\frac{m_1}{m_2} = \frac{V_2}{V_1} = \frac{27}{8}$.Assuming constant nuclear density $ho$,the mass of a nucleus is $m = ho 	imes \frac{4}{3} \pi R^3$.Therefore,$\frac{m_1}{m_2} = \frac{R_1^3}{R_2^3} = \frac{27}{8}$.Taking the cube root on both sides,$\frac{R_1}{R_2} = \left( \frac{27}{8} ight)^{1/3} = \frac{3}{2}$.Thus,the ratio of the radii is $3:2$.

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

Similar Questions

$1 \, \text{atomic mass unit (amu)}$ is equal to:

Nuclear reactions are given as $(i)$ $(n, p) {}_{16}S^{32} \to {}_{15}P^{32}$,$(ii)$ $(p, \alpha) {}_{8}O^{16} \to {}_{9}F^{19}$,$(iii)$ ${}_{7}N^{14} + ? \to {}_{6}C^{14} + {}_{1}H^{1}$. The missing particles or nuclides in these reactions are respectively:

$\pi$ mesons can be

The ratio of the nuclear radii of ${}^{135}Cs$ to ${}^{40}Ca$ is:

The nuclear force between two nucleons is explained by

Vedclass Test Series

Exam Paper Generator

Online Exam Module