Nuclear Physics and RadioactivityHard
Question
The activity of radio active sample is measured as N0 counts per minute at t = 0 and N0 /e counts per minute at t = 5 minute. The time (in minutes) at which the activity reduces to half its value is
Options
A.loge
B.
C.5 log102
D.5 loge2
Solution
According to activity law
R = R0e-λt .....(i)
where,
R0 = initial activity at t = 0
R = activity at time t
λ = decay constant
According to given problem
R0 = N0 counts per minute
R =
counts per minute
t = 5 minute
Substituting these values in equation (i), we get
= N0e-5λ
e-1 = e-5λ
5λ = 1 or λ =
per minute
At t = T1/2 the activity reduces to
where T1/2 = half life of a radioactive sample
From equation (i), we get
= R0e-λT1/2
e-λT1/2 = 2
Taking natural logarithms of both sides above equation, we get
λT1/2 = loge2
or T1/2
= 5loge 2 minutes
R = R0e-λt .....(i)
where,
R0 = initial activity at t = 0
R = activity at time t
λ = decay constant
According to given problem
R0 = N0 counts per minute
R =
counts per minutet = 5 minute
Substituting these values in equation (i), we get
= N0e-5λe-1 = e-5λ
5λ = 1 or λ =
per minuteAt t = T1/2 the activity reduces to
where T1/2 = half life of a radioactive sample
From equation (i), we get
= R0e-λT1/2 e-λT1/2 = 2
Taking natural logarithms of both sides above equation, we get
λT1/2 = loge2
or T1/2
= 5loge 2 minutesCreate a free account to view solution
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