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Question: What is the half-life of a Radium-226 if its decay rate is \(0.000436\) ? A) \(t = 1237\) B) \(t...

What is the half-life of a Radium-226 if its decay rate is 0.0004360.000436 ?
A) t=1237t = 1237
B) t=1365t = 1365
C) t=1440t = 1440
D) t=1590t = 1590

Explanation

Solution

The given decay rate of Radium-226 we will use the growth function to find the general equation of the half-life of the radium-226. Then we will substitute the given values in the formula and rearrange the terms so that we can simplify it to our advantage.

Complete step by step answer:
The given decay rate of the Radium-226 is 0.0004360.000436 .
That means Radium is decaying at this rate.
We need to find the amount of time it will take to become half of its half-life of the Radium-226.
We know that the exponential growth function is given by the following:
A=A0ertA = {A_0}{e^{ - rt}} … (1)
In the above formula first understand what each term indicates.
The term AA is the mass of the radium present at time tt and rr is the decay rate.
This implies that A0{A_0} indicates the amount of Radium-226 present at time t=0t = 0.
Therefore, we need to find the time at which the amount is half that means we need to find tt for which A=12A0A = \dfrac{1}{2}{A_0} .
Substitute this in the equation (1).
12A0=A0ert\dfrac{1}{2}{A_0} = {A_0}{e^{ - rt}}
We can cancel A0{A_0} from both sides and substitute r=0.000436r = 0.000436 .
Therefore,
12=e0.000436t\dfrac{1}{2} = {e^{0.000436t}}
Since the variable is in the exponent, we will take logarithm on both sides.
ln(12)=0.000436t\ln \left( {\dfrac{1}{2}} \right) = - 0.000436t
Note that ln(12)=0.6391\ln \left( {\dfrac{1}{2}} \right) = - 0.6391 .
Simplify the above equation for tt as follows:
t=10.000436×0.6391t = \dfrac{1}{{0.000436}} \times 0.6391
This gives us an approximate answer as follows:
t1590t \approx 1590
Thus, the correct answer is D.

Note:
Note that the given data is about the radioactive element Radium-226. The decay rate is given so we will use the decay rate and the exponential growth function to calculate the half-life. There are two types of logarithmic: one is log\log and the other is ln\ln , so correct logarithmic is used to determine the correct solution.