
Half-life of a radioactive substance is $ 20 $ minutes. The time between $ 20\% $ and $ 80\% $ decay will be:
(A) $ 40 $ minutes
(B) $ 20 $ minutes
(C) $ 25 $ minutes
(D) $ 30 $ minutes
Answer
504.6k+ views
Hint :to solve this question we have to know about radioactive substances. We know that radioactive substances are molecules that rot normally. They can emit alpha particles, beta particles and gamma radiation. Not at all like X-beam sources can’t they be killed, so their control is more troublesome. We can say, the radioactive half-life isn't equivalent to the normal lifetime.
Complete Step By Step Answer:
Here, given that, the half-life of a radioactive substance is $ 20 $ min. so, $ {t_{\dfrac{1}{2}}} = 20 $ min.
For, $ 20\% $ decay, we have $ 80\% $ of the substance left,
Hence
$ \dfrac{{80{N_0}}}{{100}} = {N_0}{e^{ - \lambda {l_{20}}}} $
Where, $ {N_0} $ is the initial decayed substance and $ {t_{20}} $ is the time taken for $ 20\% $ decay.
For $ 80\% $ decay, we have $ 20\% $ of the substance left.
Hence,
$ \dfrac{{20{N_0}}}{{100}} = {N_0}{e^{ - \lambda {t_{80}}}} $
Dividing the first equation and the second equation we will get,
$ \ln 4 = \lambda ({t_{80}} - {t_{20}}) $
Or, $ 2\ln 2 = ({t_{80}} - {t_{20}})0.693/{t_{\dfrac{1}{2}}} $
Then, $ ({t_{80}} - {t_{20}}) = 40 $ min.
So, the right answer will be option number A.
Note :
We also have to know that the unconstrained breakdown of a nuclear core of a radioactive substance bringing about the discharge of radiation from the core is known as radioactive rot. The nuclide which goes through rot in a radioactive interaction is a parent nuclide, and the nuclide which is delivered in the radioactive cycle is a little girl nuclide. We realize that, radioactive rot (otherwise called atomic rot, radioactivity, radioactive deterioration or atomic breaking down) is the interaction by which a precarious nuclear core loses energy by radiation. A material containing flimsy cores is viewed as radioactive. Three of the most well-known kinds of rot are alpha rot, beta rot, and gamma rot, all of which include transmitting at least one particle or photon. The feeble power is the instrument that is liable for beta rot, while the other two are administered by the typical electromagnetic and solid powers.
Complete Step By Step Answer:
Here, given that, the half-life of a radioactive substance is $ 20 $ min. so, $ {t_{\dfrac{1}{2}}} = 20 $ min.
For, $ 20\% $ decay, we have $ 80\% $ of the substance left,
Hence
$ \dfrac{{80{N_0}}}{{100}} = {N_0}{e^{ - \lambda {l_{20}}}} $
Where, $ {N_0} $ is the initial decayed substance and $ {t_{20}} $ is the time taken for $ 20\% $ decay.
For $ 80\% $ decay, we have $ 20\% $ of the substance left.
Hence,
$ \dfrac{{20{N_0}}}{{100}} = {N_0}{e^{ - \lambda {t_{80}}}} $
Dividing the first equation and the second equation we will get,
$ \ln 4 = \lambda ({t_{80}} - {t_{20}}) $
Or, $ 2\ln 2 = ({t_{80}} - {t_{20}})0.693/{t_{\dfrac{1}{2}}} $
Then, $ ({t_{80}} - {t_{20}}) = 40 $ min.
So, the right answer will be option number A.
Note :
We also have to know that the unconstrained breakdown of a nuclear core of a radioactive substance bringing about the discharge of radiation from the core is known as radioactive rot. The nuclide which goes through rot in a radioactive interaction is a parent nuclide, and the nuclide which is delivered in the radioactive cycle is a little girl nuclide. We realize that, radioactive rot (otherwise called atomic rot, radioactivity, radioactive deterioration or atomic breaking down) is the interaction by which a precarious nuclear core loses energy by radiation. A material containing flimsy cores is viewed as radioactive. Three of the most well-known kinds of rot are alpha rot, beta rot, and gamma rot, all of which include transmitting at least one particle or photon. The feeble power is the instrument that is liable for beta rot, while the other two are administered by the typical electromagnetic and solid powers.
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