A small quantity of the polonium isotope 210 84 Po (half-life 1.384Ý10 2 d) is p

Question

A small quantity of the polonium isotope 21084Po (half-life 1.384Ý102 d) is prepared and electroplated onto a thick sheet of copper and overcoated with a thin film of gas-tight polymer. The plating and coating are both thin enough so that none of the alpha particles resulting from the decay are absorbed, and all that exit on the film side of the copper sheet are counted. Immediately after preparation, the counter registers 7060 alpha particles per second.

a) What is the decay constant of 21084Po? D

b) How much polonium has been electroplated onto the copper?

Explanation / Answer

let decay constant be x

then N=N0e^(-xt) [where t= time]

thus,for half life we have to put N=N0/2 and t=half life=t1/2

which gives

e(-x*t1/2)=1/2

=>-x*1.384*10^2 d=-ln(2)

=>x=ln(2)/1.384*10^2 d^(-1)=0.005 d^(-1)

a)decay constant=0.005/(24*3600) /s=5.79*10^(-8)/s

b)dN/dt=N0e^(-xt)*(-x)

and dN/dt immediately after preparation(i.e. t=0)=N0(-x)=-0.005*N0

thus, N0*5.79*10^(-8)=7060

N0=7060/5.79*10^(-8)=121996800000

moles of polonium atom=121996800000/(6.022*10^23)=2.03*10^(-13)

mass of 1 mole polonium atom=210 g

so, mass of polonium atom electroplated=2.03*10^(-13)*210 g=4.26*10^(-11) g

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