Answer and show work At t o, you have 6.00x10 atoms of 24Na, which undergo beta\

Question

Answer and show work At t o, you have 6.00x10 atoms of 24Na, which undergo beta" decay. Complete the reaction: Incorrect: 2Ne Incorrect: 24Ne Incorrect: 25Ne Incorrect: 2Na Incorrect: 24Na Incorrect: 25Na Incorrect: e Correct:e Incorrect: (neither) Incorrect: v (an electron neutrino) Correct: v-bar (an electron antineutrino) +Incorrect: (neither) You are correct. You are correct. Previous Tries r receipt no. is 151-3514Previous Tries Correct: 2Mg r receipt no. is 151-819s are correct. receipt no. is 151-9509Previous Tries How much energy is released in this reaction? 88310-13 Submit Answer Unable to interpret units. Computer reads units as1 Tries 7/99 Previous Tries What is the decay constant? 0 7237Me Submit Answer Incompatible units. No conversion found between "ne and the required units.Tries 0/99 Previous Tries What is the radioactivity at t07 64285 71 NIC Submit Answer Incompatible units. No conversion found between "ic" and the required units.Tries 0/99 Previous Tries What is the radioactivity at t 1 hour? 7.3x10 4atom/s Submit Answer Unable to interpret units. Computer reads units as "sto/s Tries 0/99 Previous Tries

Explanation / Answer

given at t = 0
No = 6*10^9 atoms of 24Na which undergo beta decay

beta decay of 24 Na
24 Na = 24 Mg + energy (E)
now, mas sof 24 Mg = Mmg = 24.305 amu
mass of 24 Na = Mna = 23.99096278 amu
hence mass deficit in the reaction = dm = 0.31403722amu
hence energy rfeleased = dmc^2 = 0.31403722*1.6*10^-27*(3*10^8)^2 J = 0.00000000004522135968 J = 0.00000000004522135968/1.6*10^-19 eV
E = 282633498 eV = 282633.498 keV = 282.633498 MeV

now, decay constant = ln(2)/t'
t; = halflife = 14.9590 hours
hence
decay constat, lambda = ln(2)/t' = 0.046336465041 per hour

radioactivity at t = 0 is
Ao = lambda*No = 278018790.250663 disintegrations per second

at t = 1 hr
A = lambda*No*e^(-lambda*t) = 265430287.9235 disintegrations per second

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