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STD 12 - 13. Nuclei — NEET STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceNEETSTD 12 - 13. Nuclei4 Marks
MCQ
Two radioactive materials $A$ and $B$ have decay constant $5\lambda$ and $\lambda$ respectively.At $t=0$  they have the same number of nuclei, then the ratio of the number of nuclei of $A$ to that $B$ will be $(1/e)^2$ after a time interval
  • A
    $4λ$
  • B
    $2λ$
  • ✓
    $\frac{1}{{2\lambda }}$
  • D
    $\;\frac{1}{{4\lambda }}$
✓

Answer

Correct option: C.
$\frac{1}{{2\lambda }}$
c
Given : $\lambda_{A}=5 \lambda$, $\lambda_{B}=\lambda$

At $t=0,\left(N_{0}\right)_{A}=\left(N_{0}\right)_{B}$

$\frac{N_{A}}{N_{B}}=\left(\frac{1}{e}\right)^{2}$

According to radioactive decay, $\frac{N}{N_{0}}=e^{-\lambda t}$

$\therefore \,\frac{N_{A}}{\left(N_{0}\right)_{A}} =e^{-\lambda_{A} t} $  ..... $(i)$

$\frac{N_{B}}{\left(N_{0}\right)_{B}} =e^{-\lambda_{B} t}$  ..... $(ii)$

Divide $(i)$ by $(ii)$, we get

$\frac{N_{A}}{N_{B}}=e^{-\left(\lambda_{A}-\lambda_{B}\right) t}$  or, $\frac{N_{A}}{N_{B}}=e^{-(5 \lambda-\lambda) t}$

or, $\left(\frac{1}{e}\right)^{2}=e^{-4 \lambda t}$ or,  $\left(\frac{1}{e}\right)^{2}=\left(\frac{1}{e}\right)^{4 \lambda t}$

or, $4 \lambda t=2$ $ \Rightarrow \quad t=\frac{2}{4 \lambda}=\frac{1}{2 \lambda}$

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