In a radioactive sample $_{19}^{40} K$ nuclei either decay into stable $_{20}^{40} C a$ nuclei with decay constant $4.5 \times 1 0^{- 10}$ per year or into stable $_{18}^{40} A r$ nuclei with decay constant $0.5 \times 1 0^{- 10}$ per year. Given that in this sample all the stable $_{20}^{40} C a$ and $_{18}^{40} A r$ nuclei are produced by the $_{19}^{40} K$ nuclei only. In time $t \times 1 0^{9}$ years, if the ratio of the sum of stable $_{20}^{40} C a$ and $_{18}^{40} A r$ nuclei to the radioactive $_{19}^{40} K$ nuclei is $99$ , the value of t will be? [Given: In $10 = 2.3$ ]

Question

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Answer 1

$t = 0$

$\frac{d N}{d t} = 2 (λ_{1} + λ_{2}) \times N$
$l o g_{e} (\frac{N}{N _{0}}) = - (λ_{1} + λ_{2}) t$
$2.3 \times l o g_{10} (\frac{N _{0}}{N _{0} / 1 0 0}) = 5 \times 1 0^{- 10} t$
$\frac{2 . 3 0 3 \times 2}{5 \times 1 0 ^{- 10}} = t$
$2.303 \times 0.4 \times 1 0^{10} = t$
$t = 9.2 \times 1 0^{9}$ year.

Answer 2

t = 0

\frac{d N}{d t} = 2 (λ_{1} + λ_{2}) \times N

l o g_{e} (\frac{N}{N _{0}}) = - (λ_{1} + λ_{2}) t

2.3 \times l o g_{10} (\frac{N _{0}}{N _{0} / 1 0 0}) = 5 \times 1 0^{- 10} t

\frac{2 . 3 0 3 \times 2}{5 \times 1 0 ^{- 10}} = t

2.303 \times 0.4 \times 1 0^{10} = t

t = 9.2 \times 1 0^{9}

year.