I believe you are over thinking this.
t1/2 in day = 0.693/k
Solve for k and divide by 365 to convert to years. Compare with the list of those given.
Uranium-234 has half life of 2.48 * 10^5 years
Thorium-230 has half life of 8.0 * 10^4 years
Radium-226 has half life of 1.62 * 10^3 years
Lead-210 has half life of 20.4 years.
The rate constant for the decay of unstable nuclide X by alpha-particle emission is 1.17 * 10^-6 / day. What is identity of X? Use the following table above to assist in answering the question below. Please show step by step. This is too hard and difficult.
t1/2 in day = 0.693/k
Solve for k and divide by 365 to convert to years. Compare with the list of those given.
First, let's look at the rate constant given, which is 1.17 * 10^-6 / day. We need to find the nuclide X that matches this rate constant.
We know that the rate constant (k) for a first-order decay reaction can be related to the half-life (t1/2) using the equation:
k = 0.693 / t1/2
Now, we have the rate constant, so we can rearrange the equation to solve for the half-life:
t1/2 = 0.693 / k
Plugging in the given rate constant:
t1/2 = 0.693 / (1.17 * 10^-6 / day)
Now, let's simplify this calculation:
t1/2 ≈ 0.693 / (1.17 * 10^-6) [because dividing by a quantity is the same as multiplying by its reciprocal]
t1/2 ≈ 593076.923
Now, comparing this half-life to the values in the table, we can see that Uranium-234 has the closest half-life of 2.48 * 10^5 years.
Therefore, the identity of unstable nuclide X would be Uranium-234.
I hope this explanation lightened up the difficult topic and made it more enjoyable!
The rate constant for the decay of a nuclide is related to the half-life by the equation:
k = ln(2) / t(1/2)
Where:
- k is the rate constant
- ln(2) is the natural logarithm of 2 (approximately 0.693)
- t(1/2) is the half-life of the nuclide
Let's calculate the rate constants for each nuclide using their respective half-lives:
For Uranium-238:
t(1/2) = 4.51 * 10^9 years
k = ln(2) / (4.51 * 10^9 years)
= (0.693) / (4.51 * 10^9 years)
For Uranium-234:
t(1/2) = 2.48 * 10^5 years
k = ln(2) / (2.48 * 10^5 years)
= (0.693) / (2.48 * 10^5 years)
For Thorium-230:
t(1/2) = 8.0 * 10^4 years
k = ln(2) / (8.0 * 10^4 years)
= (0.693) / (8.0 * 10^4 years)
For Radium-226:
t(1/2) = 1.62 * 10^3 years
k = ln(2) / (1.62 * 10^3 years)
= (0.693) / (1.62 * 10^3 years)
For Lead-210:
t(1/2) = 20.4 years
k = ln(2) / (20.4 years)
= (0.693) / (20.4 years)
Now, let's compare the given rate constant to the calculated rate constants:
Given rate constant = 1.17 * 10^-6 / day
Comparing this with the calculated rate constants, we can match it with the rate constant for Radium-226:
k (Radium-226) = (0.693) / (1.62 * 10^3 years)
Therefore, the identity of nuclide X is Radium-226.
The rate constant for decay is given as 1.17 * 10^-6 / day.
Let's first convert the rate constant from days^-1 to years^-1:
1 day = 365.25 days (average number of days in a year)
1 year = 1/365.25 days
So, the rate constant becomes: 1.17 * 10^-6 / 365.25 years^-1
Now, let's compare this rate constant with the half-life values given in the table:
For Uranium-238 (half-life = 4.51 * 10^9 years):
The rate constant for alpha-particle emission is smaller than the rate constant we calculated above. Therefore, Uranium-238 is not the answer.
For Uranium-234 (half-life = 2.48 * 10^5 years):
The rate constant for alpha-particle emission is larger than the rate constant we calculated above. Therefore, Uranium-234 is a possible candidate.
For Thorium-230 (half-life = 8.0 * 10^4 years):
The rate constant for alpha-particle emission is larger than the rate constant we calculated above. Therefore, Thorium-230 is a possible candidate.
For Radium-226 (half-life = 1.62 * 10^3 years):
The rate constant for alpha-particle emission is larger than the rate constant we calculated above. Therefore, Radium-226 is a possible candidate.
For Lead-210 (half-life = 20.4 years):
The rate constant for alpha-particle emission is larger than the rate constant we calculated above. Therefore, Lead-210 is a possible candidate.
Comparing the rate constant to the half-life values, we can conclude that the nuclide X is Uranium-234.