Atoms And Nuclei Question 398
Question: In a certain hypothetical radioactive decay process, species A decays into species B and species B decays into species C according to the reactions: $ A\xrightarrow{{}}2B+particles+energy $ $ B\xrightarrow{{}}3C+particles+energy $ The decay constant for species is $ {\lambda_{1}}=1{{\sec }^{-1}} $ and that for the species B is $ {\lambda_{2}}=100{{\sec }^{-1}}. $ Initially $ 10^{4} $ moles of species of A were present while there was none of B and C. It was found that species B reaches its maximum number of moles of B.
Options:
A) 1
B) 2
C) 5
D) 4
Show Answer
Answer:
Correct Answer: B
Solution:
- $ \frac{dN_{A}}{dt}=-{\lambda_{1}}N_{A},,\frac{dN_{B}}{dt}=2{\lambda_{1}}N_{A}-{\lambda_{2}}N_{B}, $ $ N_{B}= $ maximum
$ \Rightarrow \frac{dN_{B}}{dt}=0 $
$ \Rightarrow 2{\lambda_{1}}N_{A}={\lambda_{2}}{N_{{B_{\max }}}}\Rightarrow {N_{{B_{\max }}}}=\frac{2{\lambda_{1}}}{{\lambda_{2}}}N_{A} $
$ \Rightarrow {N_{{B_{\max }}}}=\frac{2{\lambda_{1}}}{{\lambda_{2}}}N_{0}{{e}^{-{\lambda_{1^{t}=2}}}} $
