Question

What will be the radius of ${}^{64}Cu$ nucleus if the radius of ${}^{27}Al$ nucleus is $R_{Al}$ ?

• A. $R_{Cu}=\frac{4R_{Al}}{3}$
• B. $R_{Cu}=\frac{3R_{Al}}{4}$
• C. $R_{Cu}=\frac{5R_{Al}}{3}$
• D. $R_{Cu}=\frac{2R_{Al}}{3}$

Answer 1

The correct option is A $R_{Cu}=\frac{4R_{Al}}{3}$

What is formula for the nuclear radius?
$R=R_0(A{)}^{\frac{1}{3}}$
We have to apply the formula for both the steps.
Let's do it for the known first.
For Aluminum,
$R=R_0\left(A{)}^{\frac{1}{3}}\right(A=27)$
Putting the value,
$R_{Al}=3R_0$ . . . . . . .1)
For copper,
$R=R_0\left(A{)}^{\frac{1}{3}}\right(A=64)$
Putting the value,
$R_{Cu}=4R_0$ . . . . . . . 2)
Now from the first equation,
$R_0=\frac{R_{Al}}{3}$
Putting this value of $R_0$ in equation (2) we get
$R_{Cu}=\frac{4R_{Al}}{3}$