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If the radius of a nucleus with mass number 125 is 1.5 fermi, then radius of a nucleus with mass number 64 is
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Verified Answer
The correct answer is:
1.2 fermi
By Rutherford's experiment
$\begin{aligned}
R_1 & =R_0 A_1^{1 / 3} \\
R_2 & =R_0 A_2^{1 / 3}
\end{aligned}$
Given
$\begin{aligned}
R_1 & =1.5 \text { fermi } \\
A_1 & =125 \\
A_2 & =64 \\
R_2 & =?
\end{aligned}$
Now, $\frac{R_1}{R_2}=\left(\frac{A_1}{A_2}\right)^{1 / 3} \quad$ [from the Eqs. (i) and (ii)]
$\begin{aligned}
& \frac{1.5}{R_2}=\left(\frac{125}{64}\right)^{1 / 3} \\
& \Rightarrow \quad \frac{1.5}{R_2}=\frac{5}{4} \\
& \Rightarrow \quad R_2=\frac{1.5 \times 4}{5}=0.3 \times 4
\end{aligned}$
$\begin{aligned}
R_1 & =R_0 A_1^{1 / 3} \\
R_2 & =R_0 A_2^{1 / 3}
\end{aligned}$
Given
$\begin{aligned}
R_1 & =1.5 \text { fermi } \\
A_1 & =125 \\
A_2 & =64 \\
R_2 & =?
\end{aligned}$
Now, $\frac{R_1}{R_2}=\left(\frac{A_1}{A_2}\right)^{1 / 3} \quad$ [from the Eqs. (i) and (ii)]
$\begin{aligned}
& \frac{1.5}{R_2}=\left(\frac{125}{64}\right)^{1 / 3} \\
& \Rightarrow \quad \frac{1.5}{R_2}=\frac{5}{4} \\
& \Rightarrow \quad R_2=\frac{1.5 \times 4}{5}=0.3 \times 4
\end{aligned}$
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