Física·Castilla y León·2017·ExtraordinariaDatos generales del exameng0=9,80 m s−2g_0 = 9{,}80\,\text{m s}^{-2}g0=9,80m s−2G=6,67⋅10−11 N m2 kg−2G = 6{,}67 \cdot 10^{-11}\,\text{N m}^2\,\text{kg}^{-2}G=6,67⋅10−11N m2kg−2RT=6,37⋅106 mR_T = 6{,}37 \cdot 10^6\,\text{m}RT=6,37⋅106mMT=5,98⋅1024 kgM_T = 5{,}98 \cdot 10^{24}\,\text{kg}MT=5,98⋅1024kgK0=1/(4πϵ0)=9,00⋅109 N m2 C−2K_0 = 1/(4 \pi \epsilon_0) = 9{,}00 \cdot 10^9\,\text{N m}^2\,\text{C}^{-2}K0=1/(4πϵ0)=9,00⋅109N m2C−2μ0=4π⋅10−7 N A−2\mu_0 = 4 \pi \cdot 10^{-7}\,\text{N A}^{-2}μ0=4π⋅10−7N A−2e=1,60⋅10−19 Ce = 1{,}60 \cdot 10^{-19}\,\text{C}e=1,60⋅10−19Cme=9,11⋅10−31 kgm_e = 9{,}11 \cdot 10^{-31}\,\text{kg}me=9,11⋅10−31kgmp=1,67⋅10−27 kgm_p = 1{,}67 \cdot 10^{-27}\,\text{kg}mp=1,67⋅10−27kgc0=3,00⋅108 m s−1c_0 = 3{,}00 \cdot 10^8\,\text{m s}^{-1}c0=3,00⋅108m s−1h=6,63⋅10−34 J sh = 6{,}63 \cdot 10^{-34}\,\text{J s}h=6,63⋅10−34J s1 u=1,66⋅10−27 kg1\,\text{u} = 1{,}66 \cdot 10^{-27}\,\text{kg}1u=1,66⋅10−27kg1 eV=1,60⋅10−19 J1\,\text{eV} = 1{,}60 \cdot 10^{-19}\,\text{J}1eV=1,60⋅10−19JEjercicio5Opción A2 puntosa)1 pts¿Qué se entiende por dualidad onda-corpúsculo?b)1 ptsComplete y explique las siguientes desintegraciones: 1. X92238X2922238U→X90234X2902234Th+…\ce{^{238}_{92}U -> ^{234}_{90}Th} + \dotsX92238X2922238UX90234X2902234Th+… 2. X89228X2892228Ac→X90228X2902228Th+…\ce{^{228}_{89}Ac -> ^{228}_{90}Th} + \dotsX89228X2892228AcX90228X2902228Th+… 3. X…238X2…2238Rn→X84234X2842234Po+α\ce{^{238}_{\dots}Rn -> ^{234}_{84}Po} + \alphaX…238X2…2238RnX84234X2842234Po+α 4. X…212X2…2212Pb→X83212X2832212Bi+β−\ce{^{212}_{\dots}Pb -> ^{212}_{83}Bi} + \beta^-X…212X2…2212PbX83212X2832212Bi+β−
b)1 ptsComplete y explique las siguientes desintegraciones: 1. X92238X2922238U→X90234X2902234Th+…\ce{^{238}_{92}U -> ^{234}_{90}Th} + \dotsX92238X2922238UX90234X2902234Th+… 2. X89228X2892228Ac→X90228X2902228Th+…\ce{^{228}_{89}Ac -> ^{228}_{90}Th} + \dotsX89228X2892228AcX90228X2902228Th+… 3. X…238X2…2238Rn→X84234X2842234Po+α\ce{^{238}_{\dots}Rn -> ^{234}_{84}Po} + \alphaX…238X2…2238RnX84234X2842234Po+α 4. X…212X2…2212Pb→X83212X2832212Bi+β−\ce{^{212}_{\dots}Pb -> ^{212}_{83}Bi} + \beta^-X…212X2…2212PbX83212X2832212Bi+β−