The ratio of the radii of nuclei _{13}Al^{27} | vedclass

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

Question

(B) The radius of a nucleus is given by the relation $r = R_0 A^{1/3}$,where $A$ is the mass number.Thus,the ratio of radii is $\frac{r_1}{r_2} = \lef

Vedclass · Accepted Answer

$73$

Vedclass · Answer

(B) The radius of a nucleus is given by the relation $r = R_0 A^{1/3}$,where $A$ is the mass number.Thus,the ratio of radii is $\frac{r_1}{r_2} = \left(\frac{A_1}{A_2}\right)^{1/3}$.Given $\frac{r_1}{r_2} = \frac{3}{5}$ and $A_1 = 27$,we have $\frac{3}{5} = \left(\frac{27}{A_2}\right)^{1/3}$.Cubing both sides,we get $\frac{27}{125} = \frac{27}{A_2}$,which implies $A_2 = 125$.The number of neutrons $N$ is given by $N = A - Z$,where $Z$ is the atomic number.For nucleus $X$,$A = 125$ and $Z = 52$.Therefore,$N = 125 - 52 = 73$.

The ratio of the radii of nuclei $_{13}Al^{27}$ and $_{52}X^A$ is $3:5$. The number of neutrons in the nucleus of $X$ will be:

Similar Questions

The coordination number of $Cu$ is

Nuclear forces are

$A$ reaction between a proton and $_8^{18}O$ that produces $_9^{18}F$ must also liberate:

In the following nuclear reaction,the value of $X$ is: $_7N^{14} + _2He^4 \to X + _1H^1$

The electric potential at the surface of an atomic nucleus $(Z=50)$ of radius $9 \times 10^{-15} \,m$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module