Answer:
Half-life = 28.8 years
Explanation:
The radioactive decay of an atom follows the equation:
Ln[A] = -kt + ln[A]₀
Where [A] is actual amount of isotope after time t; k is decay constant and [A]₀ initial amount of isotope
Solving for k:
Ln[0.393g] = -k*10yrs + ln[0.500]
-0.2408 = -k*10yrs
0.02408yrs⁻¹ = k
Half-life, is:
Half-life = ln 2 / k
Half-life = ln 2 / 0.02408yrs⁻¹
Half-life = 28.8 years
The isotope [tex]38Sr^{90}[/tex] is one of the extremely hazardous species in the residues from nuclear power generation. The strontium in a 0.500-g sample diminishes to 0.393 g in 10.0 yrs, the half-life is - 28.78
Formula:
radioactive decay follows first-order kinetics,
k = [tex]\frac{1}{t}ln\frac{a}{x}[/tex]
Given:
a = initial amount = 0.5 g
x = amount of isomer after 10 years = 0.393 g
t= time = 10 years
Solution:
k = [tex]\frac{1}{10}ln\frac{0.5}{0.393}[/tex]
= 0.02408
k = [tex]\frac{0.693}{t/2}[/tex]
T(1/2) = half-life time
= [tex]\frac{0.693}{k}[/tex]
= [tex]\frac{0.693}{0.02408}[/tex]
= 28.78 years.
Thus, The isotope [tex]38Sr^{90}[/tex] is one of the extremely hazardous species in the residues from nuclear power generation. The strontium in a 0.500-g sample diminishes to 0.393 g in 10.0 yrs, the half-life is - 28.78
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