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


Integrating both sides


[
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


Therefore
........(1)
N= Amount of radioactive material after t unit time.
= initial amount of radioactive material
k= decay constant.
Half life:
, t= 30 years





(a)
The mass remains after t years N.

Now we put the value of
in the equation (1)

.........(2)
The mass remains of cesium after t years is
mg.
(b)
Putting
and t=120 years in equation (2)


Therefore the remains sample after 120 years is 11.25 mg.
(c)
Now putting N= 1 in equation (2)


Taking ln both sides


⇒t=224.76 (approx)
Therefore after 224.76 years only 1 mg will remain.