1.

What is the nuclear radius of Fe125, if that of Al27 is 3.6 Fermi?

Answer»

As, 

\(\frac{R_1}{R_2}\) \([\frac{A_1}{A_2}]^{1/3}\) 

\([\frac{125}{27}]^{1/3}\) 

\(\frac{5}{3}\) 

∴ R1\(\frac{5}{3}\) x R

\(\frac{5}{3}\) x 3.6

= 6.0 fermi.



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