Karen Simpson

2021-12-29

Half-life of a DNA of an extinct moa is 521 years. A 1 mg sample was estimated to be 6000 years old. Determine the initial amount of the DNA sample.

Joseph Lewis

Given half life

$t=6×{10}^{3}$ year
$\lambda =1.33×{10}^{-3}$
${N}_{t}={N}_{0}{e}^{-\lambda t}$

$⇒1={N}_{0}×3.42×{10}^{-4}$
${N}_{0}=2922.26mg$

Terry Ray

1) ${N}_{0}=2xmg$
half life $={T}_{\frac{1}{2}}=\frac{\mathrm{ln}2}{{t}_{\frac{1}{2}}}=\lambda$
$\lambda =\frac{0.693}{25}=0.02772$
${N}_{t}={N}_{0}{e}^{-\lambda y}$
${N}_{t}=0.1×20=2mg$
$2=20×{e}^{-0.02772×t}$
${e}^{-0.02772t}=0.1$
$-0.02772t=-2.3025$
$-0.02772t=-2.3025$
$t=83years$
Time required to reduce the amount to 10% of No.
2) ${t}_{\frac{1}{2}}=521$

${N}_{t}={N}_{0}{e}^{-\lambda t}$

$1={N}_{0}×3.422×{10}^{-4}$
${N}_{0}=2922.26mg$
3) Half life of carbon dating
$\frac{{N}_{t}}{{N}_{0}}={e}^{-\lambda t}$
$\lambda =\frac{0.693}{5730}=1.2094×{10}^{-4}$
$\frac{{N}_{t}}{{N}_{0}}={e}^{-7.256}=7.059×{10}^{-4}$
4) ${N}_{t}=0.14{N}_{0}$
$\lambda =\frac{0.693}{14}=0.0495$
${N}_{1}={N}_{0}{e}^{-\lambda t}$
$0.14={e}^{-0.0495×t}$
$-1.9661=-0.0495t$
$t=\text{time period}=39.719$

Vasquez