Respuesta :
Answer:
m(t) = m₀ e⁻⁰•⁰¹⁵ᵗ
Half-Life = 46.21 years.
Step-by-step explanation:
Radioactive reactions always follow a first order reaction dynamic
Let the initial mass of radioactive substance be m₀ and the mass at any time be m
(dm/dt) = -Km (Minus sign because it's a rate of reduction)
The question provides K = 0.015 from the given differential equation
(dm/dt) = -0.015m
(dm/m) = -0.015dt
∫ (dm/m) = -0.015 ∫ dt
Solving the two sides as definite integrals by integrating the left hand side from m₀ to m and the Right hand side from 0 to t.
We obtain
In (m/m₀) = -0.015t
(m/m₀) = (e^(-0.015t))
m = m₀ e^(-0.015t)) = m₀ e⁻⁰•⁰¹⁵ᵗ
m(t) = m₀ e⁻⁰•⁰¹⁵ᵗ
At half life, m(t) = (m₀/2), t = T(1/2)
(m₀/2) = = m₀ e⁻⁰•⁰¹⁵ᵗ
e⁻⁰•⁰¹⁵ᵗ = (1/2)
In e⁻⁰•⁰¹⁵ᵗ = In (1/2)
-0.015t = - In 2
t = (In 2)/0.015
t = (0.693/0.015)
t = 46.21 years
Half life = T(1/2) = t = 46.21 years.
Hope this Helps!!!
The half-life is 46.21.
Given that,
- the equation governing this is m′(t)=−0.015m(t) where t is in years.
Based on the above information, the calculation is as follows:
m'(t)=-0.015m(t)
[tex]m'(t)\div m(t)=-0.015[/tex]
Now Integrating both sides
[tex]ln(m(t))= -0.015t+m_0\\\\m(t)=m_0e^{-0.015t}[/tex]
At half lift
[tex]m(t)=(1\div2)m_0\\\\(1\div 2)m_0=m_0e^{-0.015t}\\\\1\div 2= e^{-0.015t}\\\\\ln(1\div 2)=-0.015t[/tex]
t=46.21
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