A nucleus ruptures into two nuclear parts whi | vedclass

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

Question

(B) According to the law of conservation of linear momentum,the initial momentum of the nucleus is zero,so the final momenta of the two parts must be

Vedclass · Accepted Answer

$1 : 2^{1/3}$

Vedclass · Answer

(B) According to the law of conservation of linear momentum,the initial momentum of the nucleus is zero,so the final momenta of the two parts must be equal and opposite: $m_1 v_1 = m_2 v_2$.Given the velocity ratio $\frac{v_1}{v_2} = \frac{2}{1}$,we have $\frac{m_2}{m_1} = \frac{v_1}{v_2} = \frac{2}{1}$.Since the density $\rho$ of nuclear matter is constant,the mass $m$ is proportional to the volume $V$,where $V = \frac{4}{3} \pi r^3$. Thus,$\frac{m_2}{m_1} = \frac{r_2^3}{r_1^3}$.Equating the ratios: $\frac{r_2^3}{r_1^3} = \frac{2}{1}$.Taking the cube root on both sides: $\frac{r_2}{r_1} = 2^{1/3}$.Therefore,the ratio of their radii $r_1 : r_2 = 1 : 2^{1/3}$.

$A$ nucleus ruptures into two nuclear parts which have their velocity ratio equal to $2 : 1$. What will be the ratio of their nuclear size (nuclear radius)?

Similar Questions

Nuclear reactions are given as:
$(i) \, \square (n, p) {_{15}}P^{32}$
$(ii) \, \square (p, \alpha) {_{8}}O^{16}$
$(iii) \, {_{7}}N^{14} (\square, p) {_{6}}C^{14}$
The missing particles or nuclides (in box $\square$) in these reactions are respectively:

The radius of germanium $(Ge)$ nuclide is measured to be twice the radius of ${ }_4^9 Be$. The number of nucleons in $Ge$ will be

The ratio of mass numbers of two nuclei is $1:3$. What is the ratio of their nuclear densities?

Describe the nuclear force and its characteristics.

The volume of a nucleus is directly proportional to

Vedclass Test Series

Exam Paper Generator

Online Exam Module