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Question: The radioactive element is spread over a room, its half-life is \(30\) days. Its activity is \(50\) ...

The radioactive element is spread over a room, its half-life is 3030 days. Its activity is 5050 times the permissive value. After how many days will it be safe?

Explanation

Solution

5050 times the permissive value means that to be the initial radioactive character of the element. If the half-life is provided, based on the level of decay the time to reach the permissive state can be determined.

Complete step by step answer:
Here for the radioactive element the initial level of radioactive character is 5050 times more than the permissive level. If the permissive radioactive level which is not harmful is considered as NN then the initial radioactive level for the specific radioactive element is 50N50N. The half-life of the given radioactive element is 30 days.
The initial radioactive level is considered as N0{N_0} and according to the given condition:
N0=50N{N_0} = 50N
The half-life of any element is denoted by t12{t_{\dfrac{1}{2}}}
Therefore, according to the equations relating the half-life with that of the radioactive nature of the element, there is a specific equation which can relate then both
0.693t12=2.303tlog10(N0N)\dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}} = \dfrac{{2.303}}{t}{\log _{10}}\left( {\dfrac{{{N_0}}}{N}} \right)
Here the values are provided for the decay constant and hence from there time to reach the final radioactive level can be determined with the value of tt.
Putting the value of t12{t_{\dfrac{1}{2}}} in days and also the value of N0{N_0} as given in the problem, we get
0.69330=2.303tlog10(50NN)\dfrac{{0.693}}{{30}} = \dfrac{{2.303}}{t}{\log _{10}}\left( {\dfrac{{50N}}{N}} \right)
0.69330=2.303tlog10(50)\Rightarrow \dfrac{{0.693}}{{30}} = \dfrac{{2.303}}{t}{\log _{10}}\left( {50} \right)
t=2.303×300.693log10(50)\Rightarrow t = \dfrac{{2.303 \times 30}}{{0.693}}{\log _{10}}\left( {50} \right)
t=99.697×1.69897\Rightarrow t = 99.697 \times 1.69897
t=169.38\Rightarrow t = 169.38
Therefore, the required number of days in which the level of radioactivity can reach a state of permissive level is 169.38days169.38days where the rate of radioactive character reaches the NN level. After all these days the room will be safe due to radioactive decay of the element.

Note:
Determining the rate of decay requires the constant value for radioactive decay. The rate of change from initial to final radioactive character depends on the decay constant as well as the half-life of the specific radioactive element.