Respuesta :
Answer:
Explanation:
a )
half life = 1600 years
50% in 1600 years
25% in next 1600 years
so 25% in total of 3200 years .
b )
disintegration constant = .693 / half life
= .693 / 1600
λ= 4.33 x 10⁻⁴ year⁻¹
m(t) = [tex]m_0e^{-\lambda t }[/tex]
m₀ is initial mass , λ = 4.33 x 10⁻⁴ year⁻¹
c )
m(t) after 3000 years , t = 3000
m(t) = [tex]m_0e^{-4.33\times10^{-4 }\times 3000 }[/tex]
= [tex]m_0e^{-1.3}[/tex]
= .2725 m₀
percentage of mass remaining = 27.25 %
25% radium-226 remain after 3200 years. 6.82 mg of sample remains after 3000 years. 15 mg of the sample will stay long about 1179. 19 years.
Half life of radium-226 = 1600 years
What is half-life?
half-life, in radioactivity, the interval of time required for one-half of the atomic nuclei of a radioactive sample to decay.
Calculation of each part can be done as
a )
half-life = 1600 years
50% in 1600 years
25% in next 1600 years
so 25% in total of 3200 years .
b )
Let decay function is
[tex]m(t) = m_o e^{-kt} \ \ \ \ \ ...(i)\\[/tex]
where k is decay constant
At half life,
[tex]m(t) = \frac{m_o}{2},t= 1600\ years\\So,\ \frac{m_o}{2} = m_oe^{-k1600}\\\frac{1}{2} = e^{-k1600}\\[/tex]
applying log on both sides
[tex]ln \frac{1}{2} = lne^{-1600k}\\-0.693147 = -1600 k \\k =\frac {0.693147}{1600}\\k = 4.332\times10^{-4}\\[/tex]
substitute the value of k in equation (i)
[tex]m(t) = m_oe^{-4.332\times10^{-4t}}\\m(t) = 25e^{-0.0004332t}[/tex]
c) when t = 3000 years
[tex]m(t) = 25e^{-0.0004332t}[/tex]
[tex]m(3000) = 25e^{-0.0004332\times3000}[/tex]
remain sample after 3000 years = 6.82 mg
d) when m(t) = 15 mg
[tex]15 = 25e^{-0.0004332t}\\e^{-0.0004332 t} = \frac{15}{25}\\applying\ log\ on\ both\ side\\ln \ e^{-0.0004332t}=ln(\frac{15}{25)}\\-0.0004332t = -0.510825\\t=\frac{0.510825}{0.0004332}\\t= 1179.19 \ years[/tex]
Hence, this is the required answer.
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