Complete the table for the radioactive iotope. (Round your anwer to 2 decimal place. Iotope : 239 Pu
1 answer:
If the radio active isotope has the half life of 24100 years, then the initial quantity is 2.16 grams
The half life in years = 24100
Consider the quantity of the radio active isotope remaining
y =
When t = 1000 the y = 1.2
y = C/2 when t = 1599
Substitute the values in the equation
C/2 =
Cancel the C in both side
1/2 =
Here we have to apply ln to eliminate the e terms
ln (1/2) = 24100k
k = ln(1/2) / 24100
k = -2.87× 10^-5
To find the initial value we have to substitute the value of k and y in the equation
1.2 = Ce^{1000 × -2.87× 10^-5}
C = 1.2 / e^(-0.0287)
C = 2.16 gram
Hence, the initial quantity of the radioactive isotope is 2.16 gram
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Answer:
Slope = 2
y-intercept = 1
x-intercept = -0.5
Standard Form ⇒ y - 2x = 1
Step-by-step explanation:
write the equation of the line through (-2 , -3) and (0,1)
The general form of the line is y = mx + c
Where m is the slope and c is the y-intercept
The slope m = (y₂ - y₁)/(x₂ - x₁) = (1 - [-3])/(0 - [-2]) = 4/2 = 2
∴ y = 2x + c
By substitution with the point (0,1) to find c
1 = 2 *0 + c
c = 1
∴ y = 2x + 1
Or y - 2x = 1 ⇒Standard Form
Also,
See the attached figure which represents the graph of the line y - 2x = 1
