Plutonium-238 undergoes spontaneous alpha decay, with a half-life of 86 yr. In s
ID: 994715 • Letter: P
Question
Plutonium-238 undergoes spontaneous alpha decay, with a half-life of 86 yr. In sears past, some cardiac pacemakers used the heal generated by this process to power their sealed thermoelectric batteries. Write an equation for the decay reaction. A typical pacemaker battery contained 150 mg of 238 Pu at the lime of manufacture. Calculate the rate of a-emission (decay events per second) of a new battery. Each emitted alpha particle carries a kinetic energy of 9.0 times 10^-13 J. Assuming that half of this energy was recoverable for electricity generation, calculate the electric power output in watts at the time of manufacture (1 W = 1 J/s). Calculate the electric power output of the battery ten years after the time of manufacture. Alpha particles readily capture electrons from their environment to become ordinary helium atoms. If the battery cases had a 5-mL volume, and leaked no He gas. calculate the gas pressure developed at 37 degree C (310 K) over 10 years of operation.Explanation / Answer
a) r = rate of dacay of Pu238 =) r = -dPu/dt =kPu = Kt = -dPu/Pu if de dacay have a comportament of order 1
Kt = -lnPuf/Puo =) as Puf = Puo/2 =) Kt = -LnPuo/2Puo =) Kt = -Ln0.5 as t = 86years =) K = -Ln0.5/86 =) K = 8.06x10-3years-1
r = KPu =) r = 8.06x10-3years-1 Pu
b) PM of Pu = 238g/mol or 238mg/mmol 150mg/238mg/mmol = 0.63mmol =) Pu = 0.63x10-3mol
r = 8.06x10-3years-1 Pu =) r = 8.06x10-3years-1 0.63x10-3mol = 5.08x10-3mol/years
now 5.08x10-3mol/years x years/365days x days/24hours x h/3600s =) 5.08x10-3mol/years x years/ 31536000s = 1.61x10-10mol/s
c) the rate is 1.61x10-10mol/s as 1mol = 6.02x1023particule and only de half sum electric energy we have
1.61x10-10mol/s x 6.02x1023particule/mol /2 = 4.8x1013particule/s and the energy for 1 particule = 9x10-13j =)
E = 4.8x1013particule/s x 9x10-13j = 43.2j/s = 43.2W
d) in ten years we have than Kt = -LnPuf/Puo =) e-KtPuo = Puf =)
PU = e-8.06x10-3years-1 x 10x 0.63x10-3mol = 0.58x10-3mol
r = 0.58x10-3mol x 8.06x10-3years-1 = 4.7x10-3mol/year = 4.7x10-3mol/year x years/31536000s = 1.5x10-10mol/s
particule and only de half sum electric energy we have
1.5x10-10mol/s x 6.02x1023particule/mol /2 = 4.5x1013particule/s and the energy for 1 particule = 9x10-13j =)
E = 4.5x1013particule/s x 9x10-13j = 40.5j/s = 40.5W
e) number of mol of He formed = n initial -n final = 0.63x10-3mol - 0.58x10-3mol = 0.05x10-3mol of He using the ideal gas equation we have
P = nRT/V =) P = 0.05x10-3molx0.082atml/mol k 310K /5x10-3l = 0.25atm P = 0.25atm