k = 0.693/t1/2
Then ln(No/N) = kt
No = 50 microcuries
N = 6.25 microcuries
k from above (t1/2 = 30 years).
t = solve for this.
How long will it take this sample to be reduced to a residual radioactivity of 6.25 microcuries?
Is there a formula to calculate half life?
Then ln(No/N) = kt
No = 50 microcuries
N = 6.25 microcuries
k from above (t1/2 = 30 years).
t = solve for this.
N(t) = N0 * (1/2)^(t / T)
Where:
N(t) is the final amount of the radioactive substance after time t
N0 is the initial amount of the radioactive substance
t is the time elapsed
T is the half-life of the radioactive substance
In this case, we want to find the time it takes for the Cesium 137 sample to be reduced to a residual radioactivity of 6.25 microcuries, given that its initial radioactivity is 50 microcuries and its half-life is 30 years.
Let's plug in the values into the formula and solve for t:
6.25 = 50 * (1/2)^(t / 30)
To solve for t, we need to isolate the variable. We can divide both sides of the equation by 50:
6.25 / 50 = (1/2)^(t / 30)
0.125 = (1/2)^(t / 30)
Now, to solve for t, we take the logarithm (base 1/2) on both sides:
log base (1/2) (0.125) = log base (1/2) ( (1/2)^(t / 30) )
To simplify this, we use the property of logarithms which states that log base b (b^x) = x:
log base (1/2) (0.125) = t / 30
Now, we can solve for t by multiplying both sides by 30:
30 * log base (1/2) (0.125) = t
Using a calculator or logarithmic tables, we find that log base (1/2) (0.125) is equal to 3. We substitute this value in the equation:
30 * 3 = t
t = 90 years
Therefore, it will take approximately 90 years for the Cesium 137 sample to be reduced to a residual radioactivity of 6.25 microcuries.