Física·Andalucía·2020·Variante Model DEjercicio42,5 puntosDatosc=3⋅108 m s−1c = 3 \cdot 10^8\,\text{m s}^{-1}c=3⋅108m s−11 u=1,66⋅10−27 kg1\,\text{u} = 1{,}66 \cdot 10^{-27}\,\text{kg}1u=1,66⋅10−27kgm(X24X2224He)=4,002603 um(\ce{^{4}_{2}He}) = 4{,}002603\,\text{u}m(X24X2224He)=4,002603um(X12X2122H)=2,014102 um(\ce{^{2}_{1}H}) = 2{,}014102\,\text{u}m(X12X2122H)=2,014102uaAjuste razonadamente las siguientes reacciones nucleares:X1327X213227Al+X24X2224He→X1530X215230P+XZAX2Z2AX;X1123X211223Na+X12X2122H→X1124X211224Na+XZA′X2Z2A′X\ce{^{27}_{13}Al + ^{4}_{2}He -> ^{30}_{15}P + ^{A}_{Z}X} \quad ; \quad \ce{^{23}_{11}Na + ^{2}_{1}H -> ^{24}_{11}Na + ^{A'}_{Z}X}X1327X213227Al+X24X2224HeX1530X215230P+XZAX2Z2AX;X1123X211223Na+X12X2122HX1124X211224Na+XZA′X2Z2A′X1 pt1 ptbCalcule la energía liberada en la formación de 5⋅10255 \cdot 10^{25}5⋅1025 núcleos de helio: X12X2122H+X12X2122H→X24X2224He\ce{^{2}_{1}H + ^{2}_{1}H -> ^{4}_{2}He}X12X2122H+X12X2122HX24X2224He1,5 pts1,5 pts
