a) Đúng

b) Sai. \(H = {H_0} \cdot {2^{\frac{{ - t}}{T}}} = 1,48 \cdot {10^9} \cdot {2^{\frac{{ - 10,54}}{{5,27}}}} = 0,37 \cdot {10^9}\;{\rm{Bq}}\)

c) Sai. \({H_0} = \lambda {N_0} = \frac{{\ln 2}}{T} \cdot {N_0} \Rightarrow 1,48 \cdot {10^9} = \frac{{\ln 2}}{{5,27 \cdot 365 \cdot 24 \cdot 60 \cdot 60}} \cdot {N_0} \Rightarrow {N_0} \approx 3,55 \cdot {10^{17}}\)

\({n_0} = \frac{{{N_0}}}{{{N_A}}} = \frac{{3,55 \cdot {{10}^{17}}}}{{6,02 \cdot {{10}^{23}}}} \approx 5,9 \cdot {10^{ - 7}}\;{\rm{mol}}\)

\({m_0} = {n_0}M = 5,9 \cdot {10^{ - 7}} \cdot 60 \approx 3,54 \cdot {10^{ - 5}}\;{\rm{g}} = 35,4\mu \;{\rm{g}}\)

d) Đúng. \(\Delta N = {N_0}\left( {1 - {2^{\frac{{ - t}}{T}}}} \right) = 3,55 \cdot {10^{17}} \cdot \left( {1 - {2^{\frac{{ - 4}}{{5,27365 \cdot 24}}}}} \right) \approx 2,13 \cdot {10^{13}}\)

\(Q = 0,5 \cdot \Delta N \cdot E \cdot \frac{S}{{4\pi {r^2}}} = 0,5 \cdot 2,13 \cdot {10^{13}} \cdot 2,5 \cdot 1,6 \cdot {10^{ - 13}} \cdot \frac{{1,5}}{{4\pi \cdot {4^2}}} \approx 0,0318\;{\rm{J}}\)

\(D = \frac{Q}{m} = \frac{{0,0318}}{{70}} \approx 4,5 \cdot {10^{ - 4}}Gy\)

Số proton trong hạt nhân americium \(_{95}^{241}{\rm{Am}}\) là