The activity of a radioactive sample falls from 700s-1 to 500s-1 in 30 minute. Its half-life is close to
Answers
Answered by
1
From equation of radioactivity we know
A
t
=A
0
e
−λt
Where A
t
= Amount of material at time 't'
A
0
= Amount of substance at t=0
λ= decay constant
t= time.
Taking log we have
ln[
A
t
A
0
]=λt
For half life A
t
=
2
A
0
.
⇒ln 2=λt
1/2
__(A) t
1/2
= half life
From given condition.
ln[
500
700
]=λ(30 min) __(B)
(A)−(B)
ln[7/5]
ln2
=
30
t
1/2
⇒t
1/2
=
ln[7/5]
ln2×30
⇒t
1/2
=
0.336
0.693×30
⇒t
1/2
=61.8min
Similar questions