Two radioactive substances A and B have decay constants $5 λ$ and $λ$ respectively. At $t = 0$ they have the same number of nuclei. The ratio of number of nuclei of A to those of B will be $(\frac{1}{e})^{2}$ after a time interval

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Answer 1

Decay equation is given by $N = N_{o} e^{- λ t}$
Where, $λ$ is Decay constant; $N$ is number of atom at time $t$
So, number of atom of $A$ is given by, $N_{A} = N_{o} e^{- 5 λ t}$
And that of $B$ are given by, $N_{B} = N_{o} e^{- λ t}$
Taking ration we get: $\frac{N _{A}}{N _{B}} = e^{- 4 λ t}$
Given that: $\frac{N _{A}}{N _{B}} = e^{- 2}$
So, this will happen when, $t = \frac{1}{2 λ}$

Answer 2

Decay equation is given by

N = N_{o} e^{- λ t}

Where,

λ

is Decay constant;

N

is number of atom at time

t

So, number of atom of

A

is given by,

N_{A} = N_{o} e^{- 5 λ t}

And that of

B

are given by,

N_{B} = N_{o} e^{- λ t}

Taking ration we get:

\frac{N _{A}}{N _{B}} = e^{- 4 λ t}

Given that:

\frac{N _{A}}{N _{B}} = e^{- 2}

So, this will happen when,

t = \frac{1}{2 λ}

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