Question 503: Atoms & Nuclei

Question

At any instant, two elements X
1
and X
2
have same number of radioactive atoms. If the decay constant of X
1
and X
2
are 10 $$\lambda$$ and $$\lambda$$ respectively, then the time when the ratio of their atoms becomes $${1 \over e}$$ respectively will be :

Accepted Answer

Number of radioactive nuclei at any time is
$$N = {N_0}{e^{ - \lambda t}}$$
Initial number of nuclei is same for sample X
1
& X
2
.
Let after time 't' the ratio of their atoms become $${1 \over e}$$
$$ \Rightarrow {{{N_0}{e^{ - 10\lambda  	imes t}}} \over {{N_0}{e^{ - \lambda  	imes t}}}} = {1 \over e} \Rightarrow {e^{ - 9\lambda t}} = {e^{ - 1}} \Rightarrow t = {1 \over {9\lambda }}$$
Hence, $$t = {1 \over {9\lambda }}$$

Answer

$${1 \over {5\lambda }}$$

Answer

$${1 \over {11\lambda }}$$

Answer

$${1 \over {9\lambda }}$$

Answer

$${1 \over {6\lambda }}$$

Atoms & Nuclei

At any instant, two elements X 1 and X 2 have same number of radioactive atoms. If the decay constant of X 1 and X 2 are 10 $\lambda and \lambda respectively, then the time when the ratio of their atoms becomes {1 \over e}$ respectively will be :

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