Answer:
(a) The mass remains after t years is
mg.
(b)Therefore the remains sample after 120 years is 11.25 mg.
(c)Therefore after 224.76 years only 1 mg will remain.
Step-by-step explanation:
The differential equation of decay
![\frac{dN}{dt}=-kN](https://tex.z-dn.net/?f=%5Cfrac%7BdN%7D%7Bdt%7D%3D-kN)
![\Rightarrow \frac{dN}{N}=-kdt](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cfrac%7BdN%7D%7BN%7D%3D-kdt)
Integrating both sides
![\int \frac{dN}{N}=\int-kdt](https://tex.z-dn.net/?f=%5Cint%20%5Cfrac%7BdN%7D%7BN%7D%3D%5Cint-kdt)
![\Rightarrow ln|N|=-kt+c_1](https://tex.z-dn.net/?f=%5CRightarrow%20ln%7CN%7C%3D-kt%2Bc_1)
[
is arbitrary constant ]
![[ e^{c_1}=c ]](https://tex.z-dn.net/?f=%5B%20e%5E%7Bc_1%7D%3Dc%20%5D)
Initial condition is,
when t=0
![\Rightarrow N_0=ce^{-k.0}](https://tex.z-dn.net/?f=%5CRightarrow%20N_0%3Dce%5E%7B-k.0%7D)
![\Rightarrow N_0= c](https://tex.z-dn.net/?f=%5CRightarrow%20N_0%3D%20c)
Therefore
........(1)
N= Amount of radioactive material after t unit time.
= initial amount of radioactive material
k= decay constant.
Half life:
, t= 30 years
![\therefore \frac12 N_0= N_0e^{-k\times 30}](https://tex.z-dn.net/?f=%5Ctherefore%20%5Cfrac12%20N_0%3D%20N_0e%5E%7B-k%5Ctimes%2030%7D)
![\Rightarrow \frac12=e^{-30t}](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cfrac12%3De%5E%7B-30t%7D)
![\Rightarrow -30k= ln|\frac12|](https://tex.z-dn.net/?f=%5CRightarrow%20-30k%3D%20ln%7C%5Cfrac12%7C)
![\Rightarrow k= \frac{ln|\frac12|}{-30}](https://tex.z-dn.net/?f=%5CRightarrow%20k%3D%20%5Cfrac%7Bln%7C%5Cfrac12%7C%7D%7B-30%7D)
![\Rightarrow k=\frac{ln|2|}{30}](https://tex.z-dn.net/?f=%5CRightarrow%20k%3D%5Cfrac%7Bln%7C2%7C%7D%7B30%7D)
(a)
The mass remains after t years N.
![N_0=180](https://tex.z-dn.net/?f=N_0%3D180)
Now we put the value of
in the equation (1)
![\therefore N=180 \times e^{-\frac{ln|2|}{30}t}](https://tex.z-dn.net/?f=%5Ctherefore%20N%3D180%20%5Ctimes%20e%5E%7B-%5Cfrac%7Bln%7C2%7C%7D%7B30%7Dt%7D)
.........(2)
The mass remains of cesium after t years is
mg.
(b)
Putting
and t=120 years in equation (2)
![N=180e^{-0.023\times120}](https://tex.z-dn.net/?f=N%3D180e%5E%7B-0.023%5Ctimes120%7D)
![\Rightarrow N=11.25](https://tex.z-dn.net/?f=%5CRightarrow%20N%3D11.25)
Therefore the remains sample after 120 years is 11.25 mg.
(c)
Now putting N= 1 in equation (2)
![1= 180\times e^{-0.023t}](https://tex.z-dn.net/?f=1%3D%20180%5Ctimes%20e%5E%7B-0.023t%7D)
![\Rightarrow \frac{1}{180}=e^{-0.023t}](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cfrac%7B1%7D%7B180%7D%3De%5E%7B-0.023t%7D)
Taking ln both sides
![\Rightarrow ln|\frac{1}{180}|=ln|e^{-0.023t}|](https://tex.z-dn.net/?f=%5CRightarrow%20ln%7C%5Cfrac%7B1%7D%7B180%7D%7C%3Dln%7Ce%5E%7B-0.023t%7D%7C)
![\Rightarrow -ln|180|=-0.023t](https://tex.z-dn.net/?f=%5CRightarrow%20-ln%7C180%7C%3D-0.023t)
⇒t=224.76 (approx)
Therefore after 224.76 years only 1 mg will remain.