Question

Polonium- 210 , Po210 , decays to lead- 206 , Pb206 , by alpha emission according to the equation Po84210⟶Pb82206+He24 If the half-life, ????1/2 , of Po210 is 138.4 days , calculate the mass of Pb206 produced from a 561.0 mg sample of polonium(IV) chloride, PoCl4 , that is left untouched for 333.8 days . Assume that the only polonium isotope present in the sample is the Po210 isotope. The isotopic molar masses of Po210 is 209.98 g/mol and Pb206 is 205.97 g/mol .

Answer 1

Answer:

0.269 g

Explanation:

$Po_{210} ^{84}$  ⟶  $Pb_{206} ^{82}$+  $+ He_{4} ^{2}$

Data:

Half-life of  $Po_{210} ^{84}$  (T(1/2)) = 138.4 days

Mass of PoCl4 = 561 mg (0,561 g) and molecular weight of PoCl4 = 350. 79 g/mol  

Time = 338.8 days  

Isotopic masses  

$Po_{210} ^{84}$ = 209.98 g/mol  

$Pb_{206} ^{82}$ = 205.97 g/mol  

Concepts  

Avogadro’s number: This is the number of constituent particles that are contained in a mol of any substance. These constituted particles can be atoms, molecules or ions). Its value is 6.023*10^23.  

The radioactive decay law is  

N=Noe^(-λt)

Where:  

No = number of atoms in t=0

N = number of atoms in t=t (now) in this case t=338.8 days  

λ= radioactive decay constant  

The radioactive constant is related to the half-life by the next equation  

λ= $\frac{ln 2}{t(1/2)}$

so  

λ= $\frac{ln2}{138.4 days}$  =0,005008 days^(-1)

No (Atoms of  $Po_{210} ^{84}$  in t=0)

To get No we need to calculate the number of atoms of  $Po_{210} ^{84}$   in the initial sample. We have a sample of 0,561 g of PoCl4. If we get the number of moles of PoCl4 in the sample, this will be the number of moles of  $Po_{210} ^{84}$  in the initial sample.  

This is:

$\frac{0,561 g of PoCl4}{350. 79 g of PoCl4 /mol}$ = 0,001599 mol of  PoCl4

This is the number of mol of  $Po_{210} ^{84}$ in the initial sample.

To get the number of atoms in the initial sample we use the Avogadro’s number = 6.023*10^23  

0,001599 mol of  $Po_{210} ^{84}$ * 6.023*10^23 atoms/ mol of  $Po_{210} ^{84}$ = 9.632 *10^20 atoms of  $Po_{210} ^{84}$

Atoms after 338.8 days

We use the radioactive decay law to get this value  

N=Noe^(-λt)

N=9.632*10^20 e^(-0,005008 days^(-1) * 338.8 days) =1.765*10^20

This is the number of atoms of  $Po_{210} ^{84}$ in the sample after 338.8 days has passed  

The number of atoms  $Po_{210} ^{84}$ transformed is equal to the number of atoms of $Pb_{206} ^{82}$  produced.  

The number of atoms of $Po_{210} ^{84}$ transformed is No - N  

9.632 *10^20 – 1.765 *10^20 = 7.866*10^20

So, 7.866*10^20 is the number of atoms of $Pb_{206} ^{82}$ produced  

We can get the mass with the Avogadro’s number

(7.866*10^20 atoms of $Pb_{206} ^{82}$ ) / ( 6.023*10^23 atoms of $Pb_{206} ^{82}$ / mol of $Pb_{206} ^{82}$ =  0.001306 moles of $Pb_{206} ^{82}$

This number of moles have a mass of:

(0,001306 moles of $Pb_{206} ^{82}$ )* (205.97 g of $Pb_{206} ^{82}$ /mol of $Pb_{206} ^{82}$ ) = 0.269 g