Radioactive material A has decay constant 8λ and material B has decay constant λ . Initially they have the same number of nuclei. After what time, the ratio of number of nuclei of material A to that B will be (1)/(e) ?

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Correct answer: B. (1)/(7λ)

Number of undecayed nuclei present at time

t

is

N = {N}_{0}{e}^{- \lambdat}

, here

{N}_{0}

is the number of radioactive nuclei in sample at time

t = 0

.

So for

A, {N}_{A} = {N}_{0}{e}^{- 8\lambdat}

for

B, {N}_{B} = {N}_{0}{e}^{- \lambdat}

given

\dfrac{{N}_{A}}{{N}_{B}} = \dfrac{1}{e}

Then

\dfrac{{N}_{A}}{{N}_{B}} = \dfrac{{N}_{0}{e}^{- 8\lambdat}}{{N}_{0}{e}^{- \lambdat}}

\dfrac{1}{e} = {e}^{- 7\lambdat}

- 1 = - 7\lambdat

t = \dfrac{1}{7\lambda}

Radioactive material A has decay constant 8λ and material B has...

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