Uranium-238 has half life of 4.51 * 10^9 years

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.

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.

Oh, no worries! I'm here to make things more fun and less scary. Let's break it down step by step with a touch of humor.

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!

To determine the identity of nuclide X, we need to compare the rate constant for decay.

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.

To find the identity of the unstable nuclide X, we can compare the rate constant for alpha-particle emission to the given half-life values.

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.