An isotope of sodium has a half-life of 20 hours. Suppose an initial sample of this isotope has mass 10 grams. The amount of the isotope (in grams) remaining after t hours is given by:
(Note: We're assuming exponential decay here. Use the exponential decay function)
A. 10(.965936329)^t
B. 10(1.035264924)^t
C. 10e^(-.965936329t)
D. Both A and C
E. Both B and C
Answer is A for Apex
E. Both B and C
Because exponential decay follows the formula A = A0 * e^(-kt), where A is the amount remaining after time t, A0 is the initial amount, k is the decay constant, and e is the base of the natural logarithm.
The option C presents the correct exponential decay function: 10 * e^(-.965936329t).
The option B also presents an alternative correct form of the exponential decay function by using a different approach. It follows the formula A = A0 * (1 - r)^t, where r is the decay factor. In this case, r = e^(-kt). So, we have 10 * (1.035264924)^t, which is equivalent to the exponential decay function.
The correct answer is D. Both A and C.
The exponential decay function for the amount of a substance remaining after a certain time can be written as:
N(t) = N0 * e^(-kt)
Where:
N(t) is the amount remaining after time t
N0 is the initial amount
k is the decay constant
In this case, the half-life of the sodium isotope is given as 20 hours. The decay constant (k) can be determined using the formula:
k = ln(2) / half-life
k = ln(2) / 20
Now, substituting the values into the exponential decay function:
N(t) = N0 * e^(-kt)
N(t) = 10 * e^(-(ln(2)/20)t)
N(t) = 10 * e^(-0.034657359t)
Simplifying the exponential expression, we have:
N(t) = 10 * (e^(-0.034657359))^t
By comparing this equation to the given options, we can see that option A (10(.965936329)^t) and option C (10e^(-0.034657359t)) both represent the correct exponential decay function for the amount remaining after time t.
To determine which expression represents the amount of the isotope remaining after t hours, we need to use the concept of exponential decay.
Exponential decay can be modeled using the formula:
A = A_0 * e^(-rt)
Where:
A = the amount remaining after time t
A_0 = the initial amount or mass of the substance
r = decay constant (which can be calculated as ln(2) / half-life)
t = time elapsed
In this case, the given half-life of the sodium isotope is 20 hours, so we can find the value of r as follows:
r = ln(2) / half-life = ln(2) / 20
Now, let's examine the given options:
A. 10(.965936329)^t
B. 10(1.035264924)^t
C. 10e^(-.965936329t)
D. Both A and C
E. Both B and C
Option A: 10(.965936329)^t
This expression represents the initial amount (10 grams) multiplied by a factor less than 1 raised to the power of t. However, it does not include the exponential decay function e^(-rt) and therefore does not accurately represent exponential decay. Therefore, option A is not correct.
Option B: 10(1.035264924)^t
Similar to option A, this expression does not include the exponential decay function e^(-rt) and is therefore not correct.
Option C: 10e^(-.965936329t)
This expression includes the exponential decay function e^(-rt), where r = .965936329 (calculated as ln(2) / 20). Given that r is the correct value for the decay constant, this option represents the correct equation for calculating the amount of the isotope remaining after t hours. Therefore, option C is correct.
Option D: Both A and C
This option includes both options A and C. However, since option A was determined to be incorrect, option D is not correct.
Option E: Both B and C
This option includes both options B and C. However, since option B was determined to be incorrect, option E is not correct.
Therefore, the correct answer is option C:
C. 10e^(-.965936329t)
a(t) = a(0) e^-kt
when t = 20 , a(20) = a(0) /2
so
0.5 = e*-20 k
-20 k = ln 0.5 = -.693
k = .03466
so
a(t) = a(0) e^-.03466 t