Question
Question: Some amount of a radioactive substance (half-life=10days) is spread inside a room and consequently t...
Some amount of a radioactive substance (half-life=10days) is spread inside a room and consequently the level of radiation becomes 50 times the permissible level for normal occupancy of the room. After how many days will the room be safe for occupation?
Solution
The question is based on the concept of radioactivity and half-life. The general formula of half-life of a substance is: N0N=(21)t1/2t, where(t1/2) refers to the half-life of the radioactive substance. (N) is the final amount of the radioactive substance and (N0) is the initial amount of the radioactive substance.
Step by step solution:
Let’s start by understanding all the information given in the problem. The half-life (t1/2) of the radioactive substance is given to be (10 days). Further, the initial value of the radioactive substance is 50 times the normal occupancy level, that is; N0=50× normal occupancy.
Hence, for the room to be safe again, the final amount of the radioactive substance must be N=normal occupancy. Therefore, N0N=501.
Now, we will use the half-life formula given by; N0N=(21)t1/2t. Substituting in the values of the ratio of N0N=501and the half-life of the radioactive substance as 10days, we get; N0N=(21)t1/2t⇒501=(21)10t. To solve it further, we will take the values in log on both sides.
That is; 501=(21)10t⇒log(1)−log(50)=log(21)10t⇒log(1)−log(50)=10tlog[(21)].
Therefore, 0−log50=10t[log1−log2]⇒−1.69897=10t[0−log2]⇒−1.69897=10t[−0.301]⇒0.30116.9897=t⇒t=56.45 days
Hence, the house will be safe for occupancy after 56.45 days.
Note:
Let the initial amount of the radioactive substance be(N0)and the final amount of the radioactive substance after time (t) be (N). The value of (N) will be given by the relation: N=N0e−λt→(1), where (λ) is the decay constant.
Hence, the value of the decay constant becomes; N0N=e−λt⇒NN0=eλt⇒ln(NN0)=λt⇒λ=t1ln(NN0).
If we consider the time (t=t1/2). That is the radioactive substance, reduces half of its initial value in the time then: (N=2N0) at (t=t1/2). That is; 2N0=N0e−λt1/2⇒2=eλt1/2.
That is; ln(2)=λt1/2⇒t1/2=λln(2). Therefore: λ=t1/20.693→(2)
Hence, if the value of the decay constant (λ) is small, then the half-life of the radioactive substance will be large and when the value of (λ) is large, the radioactive substance will disintegrate quickly.