The formula for a nucleus's radius is typically \(r = r_0 \cdot A^{1/3}\), where A denotes the mass number and \(r_0\) is a constant. Consequently, the ratio of the nuclear radii of two elements, \(\frac{r_A}{r_B}\), can be stated as:
\[\frac{r_A}{r_B} = \left(\frac{A_A}{A_B}\right)^{1/3}\]For elements with mass numbers \(A_A = 216\) and \(A_B = 27\), the ratio is calculated as:
\[\frac{r_A}{r_B} = \left(\frac{216}{27}\right)^{1/3} = (8)^{1/3} = 2\]Thus, the ratio of the nuclear radii is: 2 : 1
Assuming the experimental mass of \( {}^{12}_{6}\text{C} \) as 12 u, the mass defect of \( {}^{12}_{6}\text{C} \) atom is____MeV/\( c^2 \).
(Mass of proton = 1.00727 u, mass of neutron = 1.00866 u, 1 u = 931.5 MeV/\( c^2 \))
The binding energy per nucleon of \(^{209} \text{Bi}\) is _______ MeV. \[ \text{Take } m(^{209} \text{Bi}) = 208.98038 \, \text{u}, \, m_p = 1.007825 \, \text{u}, \, m_n = 1.008665 \, \text{u}, \, 1 \, \text{u} = 931 \, \text{MeV}/c^2. \]