Respuesta :
a. 1104 / 138 = 8 half lives
b after 1104 days amount of polonium in sample = 38 (1/2)^8 = 0.184 g to nearest thousandth of a gram
b after 1104 days amount of polonium in sample = 38 (1/2)^8 = 0.184 g to nearest thousandth of a gram
Answer :
(a) The number of half-lives of polonium-210 are, 8
(b) The amount left in the sample after 1104 days will be, 0.391 grams.
Explanation :
First we have to determine the amount left in the sample after 1104 days.
This is a type of radioactive decay and all radioactive decays follow first order kinetics.
To calculate the rate constant, we use the formula :
[tex]k=\frac{0.693}{t_{1/2}}[/tex]
[tex]k=\frac{0.693}{138\text{ days}}[/tex]
[tex]k=5.022\times 10^{-3}\text{ days}^{-1}[/tex]
Now we have to calculate the amount left.
Expression for rate law for first order kinetics is given by :
[tex]k=\frac{2.303}{t}\log\frac{a}{a-x}[/tex]
where,
k = rate constant
t = time taken for decay process
a = initial amount or moles of the reactant
a - x = amount or moles left after decay process
Putting values in above equation, we get:
[tex]5.022\times 10^{-3}=\frac{2.303}{138\text{ days}}\log\frac{100g}{a-x}[/tex]
[tex]a-x=0.391g[/tex]
The amount left in the sample after 1104 days will be, 0.391 grams.
Now we have to calculate the number of half-lives of polonium-210.
[tex]a=\frac{a_o}{2^n}[/tex]
where,
a = amount of reactant left after n-half lives = 0.391 g
[tex]a_o[/tex] = Initial amount of the reactant = 100 g
n = number of half lives = ?
Putting values in above equation, we get:
[tex]0.391=\frac{100}{2^n}[/tex]
[tex]256=2^n[/tex]
[tex]2^8=2^n[/tex]
[tex]n=8[/tex]
Therefore, the number of half-lives of polonium-210 are, 8
