a medical technician is working with the four samples of radionuclides listed in the table below. initially, each sample contains 15.00 μmol of the radionuclide. first, order the samples by decreasing initial radioactivity. then calculate how long it will take for the amount of radionuclide in each sample to decrease to 1/8 of the initial amount.
| sample | radionuclide | | initial radioactivity | time for amount of radionuclide to decrease to 1/8 of initial amount |
| --- | --- | --- | --- | --- |
| | symbol | half - life | | |
| a | $\ce{_{35}^{77}br}$ | 57.0 hours | select | $\square$ hours |
| b | $\ce{_{6}^{11}c}$ | 20. minutes | select | $\square$ minutes |
| c | $\ce{_{25}^{52}mn}$ | 6.0 days | select | $\square$ days |
| d | $\ce{_{16}^{35}s}$ | 87.0 days | select | $\square$ days |

Question

a medical technician is working with the four samples of radionuclides listed in the table below. initially, each sample contains 15.00 μmol of the radionuclide. first, order the samples by decreasing initial radioactivity. then calculate how long it will take for the amount of radionuclide in each sample to decrease to 1/8 of the initial amount.
| sample | radionuclide | | initial radioactivity | time for amount of radionuclide to decrease to 1/8 of initial amount |
| --- | --- | --- | --- | --- |
|  | symbol | half - life |  |  |
| a | $\ce{_{35}^{77}br}$ | 57.0 hours | select | $\square$ hours |
| b | $\ce{_{6}^{11}c}$ | 20. minutes | select | $\square$ minutes |
| c | $\ce{_{25}^{52}mn}$ | 6.0 days | select | $\square$ days |
| d | $\ce{_{16}^{35}s}$ | 87.0 days | select | $\square$ days |

Accepted Answer

# Explanation:
## Step1: Relationship between half - life and initial radioactivity
The shorter the half - life of a radionuclide, the higher its initial radioactivity.
- For sample B ($^{11}_{6}C$) with a half - life of $20$ minutes (the shortest among the four).
- For sample A ($^{77}_{35}Br$) with a half - life of $57.0$ hours.
- For sample C ($^{52}_{25}Mn$) with a half - life of $6.0$ days.
- For sample D ($^{35}_{16}S$) with a half - life of $87.0$ days (the longest among the four).

So, the order of initial radioactivity (decreasing) is $B > A > C > D$

## Step2: Formula for the amount of radionuclide decay
The formula for the amount of a radioactive substance $N = N_0(\frac{1}{2})^n$, where $N_0$ is the initial amount, $N$ is the final amount, and $n$ is the number of half - lives.

We want $N=\frac{1}{8}N_0$. Substituting into the formula $\frac{1}{8}N_0=N_0(\frac{1}{2})^n$, we get $(\frac{1}{2})^n=\frac{1}{8}$, and since $\frac{1}{8}=(\frac{1}{2})^3$, then $n = 3$

## Step3: Calculate the time for each sample
- **Sample A**:
The half - life $t_{1/2}=57.0$ hours. The time $t=n\times t_{1/2}$. Since $n = 3$, $t=3\times57.0=171$ hours
- **Sample B**:
The half - life $t_{1/2}=20$ minutes. The time $t=n\times t_{1/2}$. Since $n = 3$, $t=3\times20 = 60$ minutes
- **Sample C**:
The half - life $t_{1/2}=6.0$ days. The time $t=n\times t_{1/2}$. Since $n = 3$, $t=3\times6.0=18$ days
- **Sample D**:
The half - life $t_{1/2}=87.0$ days. The time $t=n\times t_{1/2}$. Since $n = 3$, $t=3\times87.0 = 261$ days

# Answer:
- Order of initial radioactivity (decreasing): $B > A > C > D$
- Time for sample A: $171$ hours
- Time for sample B: $60$ minutes
- Time for sample C: $18$ days
- Time for sample D: $261$ days

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QUESTION IMAGE

Question

Explanation:

Step1: Relationship between half - life and initial radioactivity

Step2: Formula for the amount of radionuclide decay

Step3: Calculate the time for each sample

Answer:

sample	radionuclide		initial radioactivity	time for amount of radionuclide to decrease to 1/8 of initial amount
a	$\ce{_{35}^{77}br}$	57.0 hours	select	$\square$ hours
b	$\ce{_{6}^{11}c}$	20. minutes	select	$\square$ minutes
c	$\ce{_{25}^{52}mn}$	6.0 days	select	$\square$ days
d	$\ce{_{16}^{35}s}$	87.0 days	select	$\square$ days