Answer: The equation to calculate the mass of remaining isotope is [tex][A]=\frac{20}{10^{-0.217t}}[/tex]
Explanation:
The equation used to calculate rate constant from given half life for first order kinetics:
[tex]t_{1/2}=\frac{0.693}{k}[/tex]
where,
[tex]t_{1/2}[/tex] = half life of the reaction = [tex]\ln 4=1.386yrs[/tex]
Putting values in above equation, we get:
[tex]k=\frac{0.693}{1.386yrs}=0.5yrs^{-1}[/tex]
Rate law expression for first order kinetics is given by the equation:
[tex]k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}[/tex]
where,
k = rate constant = [tex]0.5yr^{-1}[/tex]
t = time taken for decay process
[tex][A_o][/tex] = initial amount of the sample = 20 grams
[A] = amount left after decay process = ? grams
Putting values in above equation, we get:
[tex]0.5=\frac{2.303}{t}\log\frac{20}{[A]}[/tex]
[tex][A]=\frac{20}{10^{-0.217t}}[/tex]
Hence, the equation to calculate the mass of remaining isotope is [tex][A]=\frac{20}{10^{-0.217t}}[/tex]
The equation for the mass m(t) of the remaining isotope is [tex]m(t)=20(\frac{1}{2} )^\frac{t}{1.4}[/tex]
The half life of a substance is the amount of time it takes for a substance to decay to half of its value. It is given by:
[tex]N=N_o(\frac{1}{2} )^\frac{t}{t_\frac{1}{2} } \\\\where\ N\ is\ the\ value \ of\ the\ substance\ after\ t\ years,N_o\ is\ the\ initial\ value\\of\ substance\ and\ t\frac{1}{2} \ is \ the\ half\ life[/tex]
Given that m(t) is the mass of the remaining isotope after t years, and the initial substance is 20 g with a half life of ln(4) [1.4 years). Hence:
[tex]m(t)=20(\frac{1}{2} )^\frac{t}{1.4}[/tex]
The equation for the mass m(t) of the remaining isotope is [tex]m(t)=20(\frac{1}{2} )^\frac{t}{1.4}[/tex]
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