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Sagot :
Step-by-step explanation:
A = A02^(-t/hl) ---> ln(A/A0) = -(t/hl)ln2
solving for t,
t = -(hl)ln(A/A0)/ln2
= -(1620 yrs)×ln(8/12)/ln2
= 947.6 yrs
974.6 years It will take before it has 8 grams of radium left,
Given that,
The half-life of radium is 1620 years
Laboratory has 12 grams of radium,
We have to determine,
How long will it take before it has 8 grams of radium left.
According to the question,
Laboratory has radium = 12 grams
Left radium = 8 grams
Half life of radium [tex]_t_\frac{1}{2}[/tex] = 1620 years.
[tex]N = N_o [\frac{1}{2}]^{n} \\\\8 = 12 [\frac{1}{2}] ^{n}\\\\\frac{8}{12} } =[ \frac{1}{2} ]^{n}\\\\ log2 - log3 = n( log1 - log2)\\\\0.30-0.47 = n (0 - 0.30)\\\\-0.17 = -0.30 n\\\\n = 0.56[/tex]
It will take before it has 8 grams of radium left,
[tex]t = n \times t_\frac{1}{2} \\\\t = 0.56 \times 1620\\\\t = 974.6 \ years[/tex]
Hence, 974.6 years It will take before it has 8 grams of radium left,
For more information about Half life click the link given below.
https://brainly.com/question/24710827
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