Respuesta :
Answer:
a) The half life of the substance is 22.76 years.
b) 5.34 years for the sample to decay to 85% of its original amount
Step-by-step explanation:
The amount of the radioactive substance after t years is modeled by the following equation:
[tex]P(t) = P(0)(1-r)^{t}[/tex]
In which P(0) is the initial amount and r is the decay rate.
A sample of a radioactive substance decayed to 97% of its original amount after a year.
This means that:
[tex]P(1) = 0.97P(0)[/tex]
Then
[tex]P(t) = P(0)(1-r)^{t}[/tex]
[tex]0.97P(0) = P(0)(1-r)^{0}[/tex]
[tex]1 - r = 0.97[/tex]
So
[tex]P(t) = P(0)(0.97t)^{t}[/tex]
(a) What is the half-life of the substance?
This is t for which P(t) = 0.5P(0). So
[tex]P(t) = P(0)(0.97t)^{t}[/tex]
[tex]0.5P(0) = P(0)(0.97t)^{t}[/tex]
[tex](0.97)^{t} = 0.5[/tex]
[tex]\log{(0.97)^{t}} = \log{0.5}[/tex]
[tex]t\log{0.97} = \log{0.5}[/tex]
[tex]t = \frac{\log{0.5}}{\log{0.97}}[/tex]
[tex]t = 22.76[/tex]
The half life of the substance is 22.76 years.
(b) How long would it take the sample to decay to 85% of its original amount?
This is t for which P(t) = 0.85P(0). So
[tex]P(t) = P(0)(0.97t)^{t}[/tex]
[tex]0.85P(0) = P(0)(0.97t)^{t}[/tex]
[tex](0.97)^{t} = 0.85[/tex]
[tex]\log{(0.97)^{t}} = \log{0.85}[/tex]
[tex]t\log{0.97} = \log{0.85}[/tex]
[tex]t = \frac{\log{0.85}}{\log{0.97}}[/tex]
[tex]t = 5.34[/tex]
5.34 years for the sample to decay to 85% of its original amount
The half life of the substance is 22.76 years
Half life is given by:
[tex]N(t)=N_o(\frac{1}{2} )^\frac{t}{t_\frac{1}{2} } \\\\Where\ N(t)\ is \ the \ value\ of\ the\ substance\ after\ t\ years,N_o\ is\ the\\original\ value\ and\ t_\frac{1}{2} \ is\ the\ half\ life[/tex]
Given that t = 1 years, N(t) = 97% of normal amount = 0.97N₀. Hence:
a)
[tex]0.97N_o=N_o(\frac{1}{2} )^\frac{1}{t_\frac{1}{2} } \\\\0.97=(\frac{1}{2} )^\frac{1}{t_\frac{1}{2} }\\\\ln(0.97)=\frac{1}{t_\frac{1}{2} }ln(0.5)\\\\t_\frac{1}{2} =22.76\ years[/tex]
b)
[tex]0.85N_o=N_o(\frac{1}{2} )^\frac{t}{22.76} \\\\0.85=(\frac{1}{2} )^\frac{t}{22.76}\\\\ln(0.85)=\frac{t}{22.76}ln(0.5)\\\\t=5.3\ years[/tex]
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