The half-life for the radioactive decay of C-14 is 5730 years.How long will it take for 30% of the C-14 atoms in a sample of C-14 to decay?
Answers
Answer:
The time is
=
2948.5
y
Explanation:
The half life of
C
−
14
is
t
1
2
=
5730
y
The radioactive constant is
λ
=
ln
2
t
1
2
=
ln
2
5730
=
1.21
⋅
10
−
4
y
−
1
Apply the equation
m
(
t
)
=
m
0
e
λ
t
m
(
t
)
m
0
=
e
λ
t
λ
t
=
ln
(
m
(
t
)
m
0
)
t
=
1
λ
ln
(
m
(
t
)
m
0
)
The time is
t
=
1
1.21
⋅
10
−
4
⋅
ln
(
0.7
)
=
2948.5
y
Explanation:
It is given:
Charcoal has the initial activity, which is denoted as A_{0}=15.3A
0
=15.3 disintegrations per minute per gram.
Charcoal has the half-life, T 12=5730T12=5730 years
After a few years, the charcoal’s final activity, A = 12.3 disintegrations per minute per gram
Constant of disintegration,
\lambda=0.693 \mathrm{T} 12=0.6935370 \mathrm{y}-1λ=0.693T12=0.6935370y−1
For the action to attain 12.3 disintegrations per minute per gram, let the time taken by the sample at a time of t year.
Sample’s activity,
A=A O e-\lambda tA=AOe−λt
A=A O e-0.6935730 \times tA=AOe−0.6935730×t
\Rightarrow \mathrm{t}=1804.3 \text { years }⇒t=1804.3 years