Question 20: Atoms & Nuclei

Question

The decay constants of two radioactive substances X and Y are 4$$\lambda$$ and $$\lambda$$ respectively. At t = 0, a sample has the same number of two nuclei. The time taken for the ratio of number of nuclei to become $${1 \over {{e^3}}}$$ will be

Accepted Answer

Number of active nuclei after time t in a sample of radioactive substance is given by
N = N
0
e
$$-$$$$\lambda$$ t
where, N
0
= initial number of nuclei at t = 0 and $$\lambda$$ = decay constant.
Here, at t = 0, number of nuclei in sample X and Y are equal,
i.e., (N
0
)
X
= (N
0
)
Y
= N
0
Also, $$\lambda$$
X
= 4$$\lambda$$ and $$\lambda$$
Y
= $$\lambda$$
So, after time t, number of active nuclei in X and Y are
$${N_X} = {N_0}{e^{ - 4\lambda t}}$$ and $${N_Y} = {N_0}{e^{ - 2\lambda t}}$$
$$	herefore$$ $${{{N_X}} \over {{N_Y}}} = {e^{ - 2\lambda t}}$$
$${1 \over {{e^3}}} = {1 \over {{e^{2\lambda t}}}}$$ [Given, $${{{N_X}} \over {{N_Y}}} = {1 \over {{e^3}}}$$]
$$ \Rightarrow 2\lambda t = 3 \Rightarrow t = {3 \over {2\lambda }}$$

Answer

$${1 \over {3\lambda }}$$

Answer

$${1 \over {2\lambda }}$$

Answer

$${2 \over {3\lambda }}$$

Answer

$${3 \over {2\lambda }}$$

Atoms & Nuclei

The decay constants of two radioactive substances X and Y are 4$\lambda and \lambda respectively. At t = 0, a sample has the same number of two nuclei. The time taken for the ratio of number of nuclei to become {1 \over {{e^3}}}$ will be

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