Respuesta :
Answer:
The volcanic eruption occurred 8,634.90 years ago.
Step-by-step explanation:
Formula used:
[tex]N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}[/tex]
where,
[tex]N_o[/tex] = initial mass of isotope = x
N = mass of the parent isotope left after the time, (t) = 35% of x=0.35x
[tex]t_{\frac{1}{2}}[/tex] = half life of the isotope = 5700 year
[tex]\lambda[/tex] = rate constant
[tex]0.35x=x\times e^{-(\frac{0.693}{5700 years})\times t}[/tex]
Now put all the given values in this formula, we get
[tex]t=8,634.90 years[/tex]
The volcanic eruption occurred 8,634.90 years ago.
The half-life of the Carbon-14 isotope is about 5700 years . It occurs naturally in all organic matter.Scientists have discovered that the charcoal from a tree burned in a volcanic eruption contains 35% of the isotope level that is normally found in living trees ,The volcanic eruption occurred 8,634.90 years ago.
Given: Half life of carbon-14 isotope = 5700 years
Carbon contains isotope level = 35%
To solve this question we will use Formula,
[tex]\rm N=N_\times e^{-\lambda t}\\\lambda=\dfrac{0.693}{t_\dfrac{1}{2}}[/tex]
where,
[tex]\rm N_0= Initial\; mass\;of\;isotope\\\\N=Mass\;of\;the\;parent\;isotope\;left\;after\;the\;time\\\\\rm t_{\dfrac{1}{2}}=\rm Half\;life \;of\; isotope \\\\\lambda=\rm Rate\;of \;constant[/tex]
where,
[tex]\rm N=0.35x\\ \rm t_{\dfrac{1}{2}} =\rm 5700\; year\\ \lambda = \dfrac{0.693}{5700\rm years}[/tex]
Now, on simplifying we get
[tex]\rm 0.35x=x\times e^{-(\dfrac{0.693}{5700})\times t }\\\\t=8.634.90 years[/tex]
Therefore,The volcanic eruption occurred 8,634.90 years ago.
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