The half-life of a radioactive substance is 5 hours. A sample is tested and found to contain 0.48 of the substance. How much of the substance was present in the sample 20 hours before the sample was tested About the author Maya
Answer: [tex]X_f=0.03[/tex] Explanation: The half-life of a substance is defined by: [tex]\displaystyle \frac{X_f}{X_i}=\left(\frac{1}{2}\right)^{t/\tau}[/tex] Xf: The final number of substance left Xi: The initial number of substance t: Time elapsed [tex]\tau[/tex] (or commonly used [tex]t_{1/2}[/tex]): Half-life From here, we know: [tex]\tau=5[/tex] [tex]X_i=0.48[/tex] [tex]t=20[/tex] [tex]Xf=?[/tex] With all the information given, we can now solve for Xf: [tex]\displaystyle \frac{X_f}{0.48}=\left(\frac{1}{2}\right)^{20/5}[/tex] [tex]\displaystyle X_f=0.48 \cdot \left(\frac{1}{2} \right)^4[/tex] [tex]X_f=0.03[/tex] After 20 hours, 0.03 of the substance will be present. Reply
Answer:
[tex]X_f=0.03[/tex]
Explanation:
The half-life of a substance is defined by:
[tex]\displaystyle \frac{X_f}{X_i}=\left(\frac{1}{2}\right)^{t/\tau}[/tex]
Xf: The final number of substance left
Xi: The initial number of substance
t: Time elapsed
[tex]\tau[/tex] (or commonly used [tex]t_{1/2}[/tex]): Half-life
From here, we know:
[tex]\tau=5[/tex]
[tex]X_i=0.48[/tex]
[tex]t=20[/tex]
[tex]Xf=?[/tex]
With all the information given, we can now solve for Xf:
[tex]\displaystyle \frac{X_f}{0.48}=\left(\frac{1}{2}\right)^{20/5}[/tex]
[tex]\displaystyle X_f=0.48 \cdot \left(\frac{1}{2} \right)^4[/tex]
[tex]X_f=0.03[/tex]
After 20 hours, 0.03 of the substance will be present.