Question

What will be the radius of ${}^{64}Cu$ nucleus if the radius of ${}^{27}Al$ 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 the formula for 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, $R=R_0(A{)}^{\frac{1}{3}}.............(A=27)$

Putting the value,

$R_{Al}=3R_0.\cdots \cdots \left(1\right)$

For copper, $R=R_0\left(A{)}^{\frac{1}{3}}\right(A=64)$

Putting the value, $R_{Cu}=4R_0.\cdots \cdots \left(2\right)$

Now from the first equation,

$R_0=\frac{R_{Al}}{3}$

Putting this value of $R_0$ in equation (2) we get

$R_{Cu}=\frac{4R_{Al}}{3}$