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Question: The half-life of a radioactive substance is 20 min. The approximate time interval (\({{t}_{2}}-{{t}_...

The half-life of a radioactive substance is 20 min. The approximate time interval (t2t1{{t}_{2}}-{{t}_{1}}) between the time t2{{t}_{2}}when 2/3 of it has decayed and time t1{{t}_{1}}when 1/3 of it had decayed is ;
(A) 14 min
(B) 20 min
(C) 28 min
(D) 7 min

Explanation

Solution

Radioactivity can be defined as the ability of a substance in which the substance decays by emission of radiation. The materials which show such phenomenon are called radioactive substances. Half-life is defined as the time taken by the material in which the number of undecayed atoms becomes half. A material containing unstable nuclei by emission of radiation gained stability. Sometimes they can be harmful.

Complete step by step answer:
Given T1/2=20min{{T}_{1/2}}=20\min , the relationship between decay constant λ\lambda and half life is T1/2=0.693λ{{T}_{1/2}}=\dfrac{0.693}{\lambda }
T1/2=0.693λ\Rightarrow {{T}_{1/2}}=\dfrac{0.693}{\lambda }
λ=0.693T1/2\Rightarrow \lambda =\dfrac{0.693}{{{T}_{1/2}}}
λ=0.69320×60\Rightarrow \lambda =\dfrac{0.693}{20\times 60}
λ=0.00057s1\Rightarrow \lambda =0.00057{{s}^{-1}}--()
Now for time t2{{t}_{2}}, 23\dfrac{2}{3}of the substance has decayed, then remaining is 13\dfrac{1}{3}. Thus, using the formula,
Now using the law of radioactivity, N=N0eλtN={{N}_{0}}{{e}^{-\lambda t}}where N is the number of undecayed Nuclei at time, t and N0{{N}_{0}}are the nuclei in the starting
13N0=N0eλt2\Rightarrow \dfrac{1}{3}{{N}_{0}}={{N}_{0}}{{e}^{-\lambda {{t}_{2}}}}---(1)
Number of undecayed atoms after time t1{{t}_{1}}, 13\dfrac{1}{3}of the substance has decayed, then remaining is 23\dfrac{2}{3}.
23N0=N0eλt1\Rightarrow \dfrac{2}{3}{{N}_{0}}={{N}_{0}}{{e}^{-\lambda {{t}_{1}}}}---(2)
Dividing equation (2) by (1) we get,
2=eλ(t2t1)\Rightarrow 2={{e}^{\lambda ({{t}_{2}}-{{t}_{1}})}}
ln2=lneλ(t2t1)\Rightarrow \ln 2=\ln \\{{{e}^{\lambda ({{t}_{2}}-{{t}_{1}})}}\\}
ln2=λ(t2t1)\Rightarrow \ln 2=\lambda ({{t}_{2}}-{{t}_{1}})
λ(t2t1)=0.3010\Rightarrow \lambda ({{t}_{2}}-{{t}_{1}})=0.3010
using equation (
) we get,
(t2t1)=0.30100.00057=1200s\therefore ({{t}_{2}}-{{t}_{1}})=\dfrac{0.3010}{0.00057}=1200s
Converting in to minutes,1200/60=20min{}^{1200}/{}_{60}=20\min

So, the correct option is B.

Note: Half-life is the time for half the radioactive nuclei in any sample to undergo radioactive decay. For example, after 2 half-lives, there will be one fourth the original material remains, after three half-lives one eight the original material remains, and so on. Half-life is a convenient way to assess the rapidity of a decay. While solving such problems we have to keep in mind that while using the formula N=N0eλtN={{N}_{0}}{{e}^{-\lambda t}}, we have to take undecayed nuclei at that time.