.5 = e^-k T
ln .5 = - k T
T = -.693/-k = .693/k
b)
T = .693 / .0001074 = 6454 years
Q(t) = Q0e−kt
where Q(t) denotes the amount of the substance present at time t (measured in years), Q0 denotes the amount of the substance present initially, and k (a positive constant) is the decay constant.
(a) Find the half-life of the substance in terms of k.
(b) Suppose a radioactive substance decays according to the formula
Q(t) = 36e−0.0001074t
How long will it take for the substance to decay to half the original amount? (Round your answer to the nearest whole number.)
ln .5 = - k T
T = -.693/-k = .693/k
b)
T = .693 / .0001074 = 6454 years
Q(t) = Q0e−kt
where Q(t) denotes the amount of the substance present at time t (measured in years), Q0 denotes the amount of the substance present initially, and k (a positive constant) is the decay constant.
(a) Find the half-life of the substance in terms of k.
(b) Suppose a radioactive substance decays according to the formula
Q(t) = 36e−0.0001238t
How long will it take for the substance to decay to half the original amount? (Round your answer to the nearest whole number.)
Let's denote the original amount of the substance as Q0 and the half-life as t1/2.
When t = t1/2, Q(t) = Q0/2.
Using the given formula Q(t) = Q0e^(-kt), we can substitute these values and solve for t1/2:
Q0/2 = Q0e^(-kt1/2)
Dividing both sides by Q0:
1/2 = e^(-kt1/2)
To isolate t1/2, we can take the natural logarithm (ln) of both sides:
ln(1/2) = ln(e^(-kt1/2))
Using the property ln(e^x) = x:
ln(1/2) = -kt1/2
Next, we can isolate t1/2 by dividing by -k:
t1/2 = ln(1/2) / -k
Therefore, the half-life of the substance in terms of k is t1/2 = ln(1/2) / -k.
(b) In this case, we are given the formula Q(t) = 36e^(-0.0001074t) and we want to find the time it takes for the substance to decay to half the original amount.
We can use the formula for the half-life derived in part (a):
t1/2 = ln(1/2) / -k
Now, we need to substitute the value of k from the given formula: k = 0.0001074.
Plugging this value into the equation:
t1/2 = ln(1/2) / -0.0001074
Using ln(1/2) ≈ -0.693 (rounded to three decimal places):
t1/2 = -0.693 / -0.0001074
Calculating this expression:
t1/2 ≈ 6456.836
Rounding to the nearest whole number, we get that it will take approximately 6457 years for the substance to decay to half its original amount.
(a) We can find the half-life by substituting Q(t) = Q0/2 into the decay formula and solving for t.
Q(t) = Q0e^(-kt)
Q0/2 = Q0e^(-kt)
Dividing both sides by Q0, we get:
1/2 = e^(-kt)
Taking the natural logarithm of both sides to isolate the exponent:
ln(1/2) = ln(e^(-kt))
Using the logarithmic property ln(x^a) = a * ln(x):
ln(1/2) = -kt * ln(e)
Since the natural logarithm of e is equal to 1, we have:
ln(1/2) = -kt
Now, we solve for t:
t = -ln(1/2) / k
The half-life of the substance in terms of k is given by the equation:
Half-life = -ln(1/2) / k
(b) Let's apply the given formula Q(t) = 36e^(-0.0001074t) to find the time it takes for the substance to decay to half the original amount.
We know that the half-life is the time it takes for Q(t) to decay to Q0/2. Therefore, we substitute Q(t) = 36e^(-0.0001074t) and Q0/2 into the formula:
Q(t) = Q0e^(-kt)
Q0/2 = 36e^(-0.0001074t)
Dividing both sides by 36e^(-0.0001074t), we get:
1/2 = e^(-0.0001074t)
Taking the natural logarithm of both sides to isolate the exponent:
ln(1/2) = ln(e^(-0.0001074t))
ln(1/2) = -0.0001074t * ln(e)
Since the natural logarithm of e is equal to 1, we have:
ln(1/2) = -0.0001074t
Now, we solve for t:
t = -ln(1/2) / 0.0001074
Plugging this into a calculator, we find:
t ≈ 6457
Therefore, it will take approximately 6457 years for the substance to decay to half the original amount.