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  

Answer 2

From the calculation, the mass of the lead isotope is 105.4 mg.

What is half life?

The term half life refers to the  time it takes to reduce the number of radioactive atoms present to half of the original number.

We know that the half life of the isotope is 138.4 days, hence;

N/No = (1/2)^t/t1/2

No =  561.0 mg

N = ?

t1/2 =  138.4 days

t =  333.8 days

Substituting values;

N/561.0 = (1/2)^ 333.8 / 138.4

N = (1/2)^ 333.8 / 138.4 * 561.0

N = 105.4 mg

Learn more about half life: https://brainly.com/question/4219437?