M(t)=Ae^(-0.421t). Find the half life
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Answer:
197.3 DAYS
Step-by-step explanation:
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Class 12
>>Applied Mathematics
>>Differential Equations
>>Applications of Differential Equations
>>Find the half life of a radioactive elem
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Find the half life of a radioactive element, if its activity decreases for 1 month by 10%.
Hard
Solution
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Correct option is B)
Activity of an isotope is measured by the number of nuclei decaying for a time unit.
Suppose that dN
d
nuclei decay for a short period of time dt.
Then the isotope activity A is expressed by the formula A=
dt
dN
d
It follows from the radioactive decay law that
N(t)=N
0
e
−λt
where N(t) is the quantity of the remaining nuclei.
Therefore,
N
d
(t)=N
0
−N(t)=N
0
−N
0
e
−λt
=N
0
(1−e
−λt
)
By differentiating with respect to t, we find the expression for activity:
A(t)=
dt
dN
d
=N
0
λe
−λt
The initial isotope activity is equal to
A(t=0)=A
0
=N
0
λ
Hence, A(t)=A
0
e
−λt
As it can be seen, the activity decreases over time by the same law as the amount of undecayed material. Substituting the expression for the half life T=
λ
ln2
in the last formula, we can write:
A(t)=A
0
e
−
T
tln2
The value of T in the last expression can be found by
e
−
T
tln2
=
A
0
A
⇒−
T
tln2
=ln
A
0
A
⇒
T
tln2
=ln
A
A
0
⇒T=
ln
A
A
0
tln2
In our case, the half life period of the given isotope is
T=
ln
A
A
0
tln2
=
ln
90
100
30ln2
≈
ln1.11
30⋅0.93
≈197.3days