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What is the nuclear radius of Fe125, if that of Al27 is 3.6 Fermi? |
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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 R2 = \(\frac{5}{3}\) x 3.6 = 6.0 fermi. |
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