Answer:
(a) [tex]N=19\times e^{-\lambda t}[/tex]
(b) 15 hours
Explanation:
half life, T = 12 hours
No = 19 g
(a) Let N be the amount remaining after time t.
Let λ be the decay constant.
[tex]\lambda =\frac {0.6931}{T}[/tex]
The equation of radioactivity used here is given by
[tex]N=N_{o}e^{-\lambda t}[/tex]
[tex]N=19\times e^{-\lambda t}[/tex]
(b) N = 8 gram
Substitute the values in above equation
[tex]\lambda =\frac {0.6931}{12}[/tex]
λ = 0.0577 per hour
So, [tex]8=19\times e^{-0.577t}[/tex]
[tex]e^{-0.0577t}=0.421[/tex]
Take natural log on both the sides
- 0.0577 t = - 0.865
t = 15 hours
Answer:
The amount of substance remaining as a function of time t is[tex]N=19 \times e^{ 0.0580 \times t}[/tex].
The time remaining for 8 grams is 15 hours.
Explanation:
Given data:
The half life is, [tex]t_{1/2}=12 \;\rm hrs[/tex].
Initial amount is, [tex]N_{0} =19 \;\rm g[/tex].
(a)
The amount of substance remaining as a function of time t is,
[tex]N=N_{0}e^{ - \lambda \times t}[/tex]
Here, [tex]\lambda[/tex] is the time constant and its value is,
[tex]\lambda= \dfrac{0.6931}{t_{1/2}} \\\lambda= \dfrac{0.6931}{12}} \\\lambda= 0.0580[/tex]
Then,
[tex]N=19 \times e^{-0.0580 \times t}[/tex]
Thus, the amount of substance remaining as a function of time t is [tex]N=19 \times e^{ 0.0580 \times t}[/tex].
(b)
Time remaining for 8 grams is,
[tex]N=19 \times e^{-0.0580 \times t}\\8=19 \times e^{-0.0580 \times t}\\\dfrac{8}{19}= e^{-0.0580 \times t}\\log(\frac{8}{19})=log(e^{-0.0580 \times t})\\-0.375=-0.0580t\\t = 15 \;\rm hrs[/tex]
Thus, the time remaining for 8 grams is 15 hours.
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