Respuesta :
Answer:
[tex]9.31\times 10^{17}\ Bq[/tex]
[tex]2.52\times 10^{7}\ Ci[/tex]
Explanation:
Calculation of the moles of Bromine-88 as:-
Mass = 3.2 mg
Also, 1 mg = 0.001 g
So, Mass = [tex]0.0032\ g[/tex]
Molar mass of Bromine-88 = 88 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{0.0032\ g}{88\ g/mol}[/tex]
1 mole of Bromine-88 contains [tex]6.023\times 10^{23}[/tex] atoms of Bromine-88
So,
[tex]\frac{0.0032}{88}[/tex] mole of Bromine-88 contains [tex]\frac{0.0032}{88}\times 6.023\times 10^{23}[/tex] atoms of Bromine-88
Atoms of Bromine-88 in the sample = [tex]2.19\times 10^{19}[/tex]
Given that:
Half life = [tex]16.3[/tex] s
The expression for half-life is:-
[tex]t_{1/2}=\frac {ln\ 2}{k}[/tex]
Where, k is rate constant
So,
[tex]k=\frac {ln\ 2}{t_{1/2}}[/tex]
[tex]k=\frac{\ln2}{16.3}\ s^{-1}[/tex]
The rate constant, k = 0.04252 s⁻¹
Disintegration is:-
Disintegrations per second = Rate constant*Number of atoms = [tex]0.04252\times 2.19\times 10^{19}\ Bq[/tex] = [tex]9.31\times 10^{17}\ Bq[/tex]
Considering that:-
1 Bq = [tex]2.7\times 10^{-11}[/tex] Ci
Thus, disintegrations = [tex]9.31\times 10^{17}\times 2.7\times 10^{-11}\ Ci[/tex] = [tex]2.52\times 10^{7}\ Ci[/tex]
