Find the mass density of $_{6}^{12} C$ nucleus. Take atomic mass of $_{6}^{12} C$ as 12.000 amu Take $R_{0} = 1.2 \times 10^{- 15} m$

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Solution

The density of the nucleus, $n = \frac{A}{(4 / 3) π R^{3}} = \frac{A}{(4 / 3) π (R_{0} A^{1 / 3})^{3}}$
$n = \frac{3}{4 π R_{0}^{3}} = \frac{3}{4 π (1.2 \times 10^{- 15})^{3}} = 1.38 \times 10^{44} m^{- 3}$

Mass density $= n M = 1.38 \times 10^{44} \times (12.00 \times 1.67 \times 10^{- 27}) = 27.65 \times 10^{17} k g / m^{3}$

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_{6}^{12} C

56

(Take R_{0} = 1.2 \times 10^{- 15} m : (56)^{\frac{1}{3}} = 3.82)

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R_{0} =

R_{0} = 1.2 \times 10^{- 15}

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R_{0} = 1.2 \times 10^{- 15}

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