Answer:
[tex]m(t)=20e^{-0.5t}[/tex]
Explanation:
Given:
Initial mass of isotope (m₀) = 20 g
Half life of the isotope [tex](t_{1/2})[/tex] = (ln 4) years
The general form for the radioactive decay of a radioactive isotope is given as:
[tex]m(t)=m_0e^{-kt}[/tex]
Where,
[tex]m(t)\to mass\ after\ 't'\ years\\t\to years\ passed\\k\to rate\ of\ decay\ per\ year[/tex]
So, the equation is: [tex]m(t)=20e^{-kt}[/tex]
At half-life, the mass is reduced to half of the initial value.
So, at [tex]t=t_{1/2},m(t)=\frac{m_0}{2}[/tex]. Plug in these values and solve for 'k'. This gives,
[tex]\frac{m_0}{2}=m_0e^{-k\times\ln 4}\\\\0.5=e^{-k\times\ln 4}\\\\Taking\ natural\ log\ on\ both\ sides,we\ get:\\\\\ln(0.5)=-k\times \ln 4\\\\k=\frac{\ln 0.5}{-\ln 4}=0.5[/tex]
Hence, the equation for the mass remaining is given as:
[tex]m(t)=20e^{-0.5t}[/tex]
