First slide

Radio activity

Question

Radioactive material 'A' has decay constant '8 $λ$ ' and material 'B' has decay constant ' $λ$ '. Initially they have same number of nuclei. After what time, the ratio of number of nuclei of material 'B' to that 'A will be e ?

Moderate

A

$\frac{1}{7 λ}$
B

$\frac{1}{8 λ}$
C

$\frac{1}{9 λ}$
D

$\frac{1}{λ}$

Solution

The number of radioactive nuclei ‘N’ at any time t is given as $N (t) = N_{0} e^{- λ t}$
where $N_{0}$ is number of radioactive nuclei in the sample at some arbitrary time t = 0 and $λ$ is the radioactive decay constant.
$Given: λ_{A} = 8 λ, λ_{B} = λ, N_{0 A} = N_{0 B} = N_{0}$
$∴ \frac{N_{B}}{N_{A}} = \frac{e^{- λ t}}{e^{- 8 λ t}}$
$e = e^{- λ t} e^{8 λ t} = e^{7 λ t} \Rightarrow 1 = 7 λ t or t = \frac{1}{7 λ}$

NOTE:-
in original question paper the question was "Radioactive material 'A' has decay constant '8 $λ$ ' and material 'B' has decay constant ' $λ$ '. Initially they have same number of nuclei. After what time, the ratio of number of nuclei of material 'B' to that 'A will be $\frac{1}{e}$ ? "
with this data value for t we get after solving we get t = $- \frac{1}{7 λ}$
Negative value of time is not possible. So given ratio in question should be $\frac{N_{B}}{N_{A}} = e$

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