Độ bền vững của hạt nhân càng cao khi

\(n = \frac{m}{M} = \frac{{0,5 \cdot {{10}^3}}}{{235}} = \frac{{100}}{{47}}\;{\rm{mol}}\)

\(N = n{N_A} = \frac{{100}}{{47}} \cdot 6,02 \cdot {10^{23}} \approx 1,{281.10^{24}}\)

\(Q = N\Delta E = 1,281 \cdot {10^{24}} \cdot 200 \cdot 1,6 \cdot {10^{ - 13}} \approx 4,1 \cdot {10^{13}}J\)

\(A = HQ = 0,2.4,{1.10^{13}} = 8,{2.10^{12}}J\)

\(t = \frac{Q}{P} = \frac{{8,{{2.10}^{12}}}}{{{{160.10}^3}}} = {5125.10^4}s \approx 593\) ngày. Chọn B

\(S = \pi {r^2} + 2\pi rh = \pi \cdot {4^2} + 2\pi \cdot 4 \cdot 10 = 96\pi \left( {\;{\rm{c}}{{\rm{m}}^2}} \right)\)

\(V = \pi {r^2}h = \pi \cdot {4^2} \cdot 10 = 160\pi \left( {\;{\rm{c}}{{\rm{m}}^3}} \right)\)

\(m = VD = 160\pi \cdot {10^{ - 6}} \cdot 1000 = 0,16\pi (\;{\rm{kg}})\)

\( - mc \cdot d{t_t} = \frac{{kS\left( {{t_t} - {t_p}} \right)}}{d} \cdot d\tau \Rightarrow - \frac{1}{{{t_t} - {t_p}}} \cdot d{t_t} = \frac{{kS}}{{mcd}} \cdot d\tau \Rightarrow - \int_{100}^4 {\frac{1}{{{t_t} - 25}}} \cdot d{t_t} = \int_0^2 {\frac{{96\pi \cdot {{10}^{ - 4}}}}{{0,16\pi \cdot 4180 \cdot 5 \cdot {{10}^{ - 3}}}}} \cdot d\tau \)

Casio shift solve \( \Rightarrow {t_t} \approx 99,{5706^\circ }{\rm{C}}\)

\(Q = mc\Delta T = 0,16\pi \cdot 4180.(100 - 99,5706) \approx 902J\)

Trả lời ngắn: 902