Respuesta :
Answer:
(a). The decay constant is [tex]1.55\times10^{-5}\ s^{-1}[/tex]
The half life is 11.3 hr.
(b). The value of N₀ is [tex]2.38\times10^{11}\ nuclei[/tex]
(c). The sample's activity is 1.87 mCi.
Explanation:
Given that,
Activity [tex]R_{0}=10\ mCi[/tex]
Time [tex]t_{1}=4\ hours[/tex]
Activity R= 8 mCi
(a). We need to calculate the decay constant
Using formula of activity
[tex]R=R_{0}e^{-\lambda t}[/tex]
[tex]\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})[/tex]
Put the value into the formula
[tex]\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})[/tex]
[tex]\lambda=0.0000154\ s^{-1}[/tex]
[tex]\lambda=1.55\times10^{-5}\ s^{-1}[/tex]
We need to calculate the half life
Using formula of half life
[tex]T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}[/tex]
Put the value into the formula
[tex]T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}[/tex]
[tex]T_{\dfrac{1}{2}}=44.719\times10^{3}\ s[/tex]
[tex]T_{\dfrac{1}{2}}=11.3\ hr[/tex]
(b). We need to calculate the value of N₀
Using formula of [tex]N_{0}[/tex]
[tex]N_{0}=\dfrac{3.70\times10^{6}}{\lambda}[/tex]
Put the value into the formula
[tex]N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}[/tex]
[tex]N_{0}=2.38\times10^{11}\ nuclei[/tex]
(c). We need to calculate the sample's activity
Using formula of activity
[tex]R=R_{0}e^{-\lambda\times t}[/tex]
Put the value intyo the formula
[tex]R=10e^{-(1.55\times10^{-5}\times30\times3600)}[/tex]
[tex]R=1.87\ mCi[/tex]
Hence, (a). The decay constant is [tex]1.55\times10^{-5}\ s^{-1}[/tex]
The half life is 11.3 hr.
(b). The value of N₀ is [tex]2.38\times10^{11}\ nuclei[/tex]
(c). The sample's activity is 1.87 mCi.
