Answer:
27.76 grams will be present in 500 years
Step-by-step explanation:
The given formula is [tex]A=A_{o}e^{kt}[/tex] , where A is the value of the substance in t years, and [tex]A_{o}[/tex] is the initial value
∵ The half-life is a substance is 375 years
- Substitute A by [tex]\frac{1}{2}A_{o}[/tex] and t by 375 to find the value of k
∴ [tex]\frac{1}{2}A_{o}=A_{o}e^{375k}[/tex]
- Divide both sides by [tex]A_{o}[/tex]
∴ [tex]\frac{1}{2}=e^{375k}[/tex]
- Insert ㏑ in both sides
∴ ㏑( [tex]\frac{1}{2}[/tex] ) = ㏑ ( [tex]e^{375k}[/tex] )
- Remember ㏑ ( [tex]e^{n}[/tex] ) = n
∵ ㏑ ( [tex]e^{375k}[/tex] ) = 375 k
∴ ㏑( [tex]\frac{1}{2}[/tex] ) = 375 k
- Divide both sides by 375
∴ k ≈ -0.00185
∴ [tex]A=A_{o}e^{-0.00185t}[/tex]
∵ 70 grams is present now
- That means the initial value is 70 grams
∴ [tex]A_{o}[/tex] = 70
∵ The time is 500 years
∴ t = 500
- Substitute the values of [tex]A_{o}[/tex] and t in the formula
∵ [tex]A=70e^{-0.00185(500)}[/tex]
∴ A = 27.76
∴ 27.76 grams will be present in 500 years
