Question
A radioactive nucleus decay to a nucleus with a decay with a decay Concept further decay to a stable nucleus Z with a decay constant initialy, there are only X nuclei and their number is set up the rate equations for the population of The population of nucleus as a function of time is givenby N_(y)X and Y\displaystyle{a}{t}{t}\hat{\in}{s}{\tan{{t}}}.
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Answer: A::B::C::DSolution: X overset(T_(1//2) = 10sec) underset(lambda_(x) = 0.1s^(-1))(rarr) Y overset(T_(1//2) = 30sec) underset(lambda_(y) = (1)/(30)s^(-1))(rarr)Z(dN_(x))/(dt) = - lambda_(x)N_(x) (dN_(y))/(dt) = - lambda_(y)N_(y)+ lambda_(x)N_(x) (dN_(z))/(dt) = - lambda_(y)N_(y) rArrN_(x_ = N_(0)e^(-lambda _(x)t)N_(y) = (lambda_(x) N_(0))/(lambda_(x) - lambda_(y)) [ e^(- lambda_(y) t) - e^(lambda_(x)t)]N_(p)(dN_(p))/(dt) = 0 - lambda_(y) N_(y) +lambda_(x) N_(x) = 0 rArr ambda_(x) N_(x) = lambda_(y) N_(y)rArr lambda_(x) (N_(0)e^(-lambda_(x) t) = lambda_(y)[(lambda_(x) N_(0))/(lambda_(x) - lambda_(y)) (e^(- lambda_(y)t - e^(lambda_(x) t)]rArr (lambda_(x) - lambda_(y))/(lambda_(y))= (e^(- lambda_(x) t))/(- e^(-lambda_(x)t)) rArr (lambda_(x))/(lambda_(y)) = e^(lambda_(x) - lambda_(y)) t)rArr log _(e) (lambda_(x))/(lambda_(y)) = (lambda_(x) - lambda_(y)) tt = (log _(e)(lambda_(x) //lambda_(y))/(lambda_(x) - lambda_(y)) = log _(e)[0.1//((1)/(30))]/(0.1 - (1)/(30) = 15log _(e) 3 :. N_(z) = N_(0)e^(-0.1(15 log_(e) 3) = N_(0) e^(log_(e) (T^(-15)) rArr N_(x) = N_(0) 3^(-15) = (10^(20))/(3sqrt(3))(dN_(y))/(dt) = 0 at t = 15 log_(e) , 3 , :. N_(p) =(lambda_(x)N_(x))/(lambda_(y)) = (10^(20))/(sqrt(3)) and N_(x) = N_(0) - N_(x) - N_(y)= 10^(20) - ((10^(20))/(3 sqrt(3))) - (10^(20))/(sqrt(3)) = 10^(20) ((3 sqrt(3)-4)/(3sqrt(3)))
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Question Text | A radioactive nucleus decay to a nucleus with a decay with a decay Concept further decay to a stable nucleus Z with a decay constant initialy, there are only X nuclei and their number is set up the rate equations for the population of The population of nucleus as a function of time is givenby N_(y)X and Y\displaystyle{a}{t}{t}\hat{\in}{s}{\tan{{t}}}. |
Answer Type | Text solution:1 |
Upvotes | 150 |