Correct Answer - Option 3 : 3000 years
Content:
Radioactive Decay: It is the process by which an unstable atomic nucleus loses energy by radiation. A material containing unstable nuclei is called radioactive material. Alpha decay, Beta decay, and Gamma decay are the most common radioactive decay. It can also be defined as the disintegration of a nucleus into its constituents. The number of radioactive disintegration is proportional to the number of atoms left that have not decayed. If N is the number of undecayed nuclei then
\(-\dfrac{dN}{dt}\propto N\)
\(\dfrac{dN}{dt}=-\lambda N\) , λ is the decay constant
\(\dfrac{dN}{N}=-\lambda dt\)
Integrating both sides-
\(\int_{N_0}^{N} \dfrac{dN}{N}= -\int_{0}^{t} \lambda dt\)
\(\ln \dfrac{N}{N_0}=-\lambda t\)
\(N=N_0 e^{-\lambda t}\)
Here, N0 is the number of nuclei initially present and N is the number of nuclei at any time t
Half lifetime(t1/2): The time elapsed before half the active nuclei decay is called the half-life and is denoted by t1/2. Suppose there are N0 active nuclei at t=0. The half-life is the time elapsed before N0/2 nuclei have decayed and N0/2 remain active.
Mathematically,
\(\dfrac{N_0}{2}=N_0e^{-\lambda t_{1/2}}\)
\(\large e^{\lambda t_{1/2}}=2\)
\(\lambda t_{1/2} = ln 2\)
\(t_{1/2}= \dfrac{ln\ 2}{\lambda}=\dfrac{0.693}{\lambda}\)
The number of nuclei left after n lifetimes is given by
\(N=N_0 \left(\dfrac{1}{2}\right)^n \)
The number of nuclei left at any time t can also be written as
\(\dfrac{N}{N_0}=\left(\dfrac{1}{2}\right)^{t/t_{1/2}}\)
In terms of Activity,
\(\dfrac{A}{A_0}=\left(\dfrac{1}{2}\right)^{t/t_{1/2}}\)
Mean lifetime: It is the sum of the average lifetime of all the nuclei present in a given radioactive sample initially. In other words, the time taken by a radioactive sample to get reduced to 1/e times the nuclei present initially is known as mean lifetime.
let mean lifetime be t= τ
then at t= τ , N= N0/e
i.e., \(\dfrac{N_0}{e}= N_0 e^{-\lambda \tau}\)
\(\dfrac{1}{e}=e^{-\lambda \tau}\)
\(\tau = \dfrac{1}{\lambda}\)
Calculation:
We know, \(\dfrac{N}{N_0}=\left(\dfrac{1}{2}\right)^{t/t_{1/2}}\)--------------------(1)
given N0 = 200 g
and N = 50 g
t1/2 =1500 years
t=?
putting these in (1)
\(\dfrac{50}{200}=\left(\dfrac{1}{2}\right)^{t/1500}\)
\(\dfrac{1}{4}=\left(\dfrac{1}{2}\right)^{t/1500}\)
\(\left(\dfrac{1}{2}\right)^2=\left(\dfrac{1}{2}\right)^{t/1500}\)
On comparing both sides
\(2=\dfrac{t}{1500}\)
t=3000 years
The correct answer is option (3)
