If the nucleus $^{27}_{13}\mathrm{Al}$ has a nuclear radius of about 3.6 fermis, then $^{125}_{52}\mathrm{Te}$

Question

If the nucleus $^{27}_{13}\mathrm{Al}$ has a nuclear radius of about 3.6 fermis, then $^{125}_{52}\mathrm{Te}$ would have its radius approximately as:

Accepted Answer

The nuclear radius $R$ is related to the mass number $A$ by the relationship $R = R_0 A^{1/3}$, where $R_0$ is a constant (approximately 1.2 fm).

For Aluminum ($Al$), the mass number $A_1 = 27$ and radius $R_1 = 3.6$ fm. 
For Tellurium ($Te$), the mass number $A_2 = 125$.

Taking the ratio: 
$\frac{R_2}{R_1} = \left(\frac{A_2}{A_1}ight)^{1/3}$ 
$\frac{R_2}{3.6} = \left(\frac{125}{27}ight)^{1/3}$ 
$\frac{R_2}{3.6} = \frac{5}{3}$ 
$R_2 = 3.6 	imes \frac{5}{3} = 1.2 	imes 5 = 6.0$ fm.

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