Question

The nuclear radius of a certain nucleus is $7.2fm$ and it has charge of $1.28\times {10}^{-17}C$ . The number of neutrons inside the nucleus is

• A. 136
• B. 142
• C. 140
• D. 132
• E. 126

Answer 1

Answer: (A)

$R=R_0{A}^{1/3}$
Here, $R=7.2\times {10}^{-15}m,R_0=1.2\times {10}^{-15}m$
$\therefore$ $A={\left(\frac{R}{R_0}\right)}^3=\left(\frac{7.2\times {10}^{-15}}{1.2\times {10}^{-15}}\right)={(6)}^3=216$
Also, atomic number $Z=\frac{q}{e}=\frac{1.28\times {10}^{-17}}{1.6\times {10}^{-19}}=80$ Therefore, number of neutrons $N=A-Z=216-80=136$