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B9. Radioactive iodine is used to treat tumors of the thyroid gland. It decays by emitting beta particles and gamma radiation. The beta emitting process is represented by the following equation:
[tex]\[{}_{53}^{131} I \rightarrow {}_Z^A Xe + {}_{-1}^0 e\][/tex]

(a) What is the nucleon number of the new nucleus [tex]\[Xe\][/tex] formed?
Nucleon number: [1]

(b) The half-life of iodine-131 is 8 days. The total dose of iodine given to a patient initially emits [tex]\[4 \times 10^8\][/tex] rays per second.

(c) How many gamma rays does the total dose of iodine emit each second after 24 days?


Sagot :

Let's break down and solve each part of the problem step-by-step:

### Part (a) Nucleon Number of the New Nucleus [tex]\(X_e\)[/tex]

Given radioactive decay:

[tex]\[ {}_{53}^{131} I \rightarrow {}_{Z}^{A} X_e + {}_{-1}^{0} e \][/tex]

In beta decay, a neutron is converted into a proton and an electron. Since the nucleon number (A) represents the total number of protons and neutrons, it remains unchanged during the emission of a beta particle. Therefore:

- The initial nucleon number of iodine-131 ([tex]\( {}_{53}^{131} I \)[/tex]) is 131.
- The new nucleus [tex]\(X_e\)[/tex] will have the same nucleon number because nucleon number remains constant during beta decay.

Thus, the nucleon number of the new nucleus [tex]\(X_e\)[/tex] is:
[tex]\[ 131 \][/tex]

### Part (b) Half-Life of Iodine-131

The problem statement provides that the half-life of iodine-131 is 8 days.

### Part (c) Gamma Rays Emission Rate After 24 Days

Given:
- The initial emission rate is [tex]\(4 \times 10^8\)[/tex] rays per second.
- The time elapsed is 24 days.
- The half-life of iodine-131 is 8 days.

To find out how many gamma rays the iodine emits after 24 days, we need to determine how many half-lives have passed in 24 days:

[tex]\[ \text{Number of half-lives} = \frac{\text{Time elapsed}}{\text{Half-life}} \][/tex]

[tex]\[ \text{Number of half-lives} = \frac{24 \text{ days}}{8 \text{ days}} = 3 \][/tex]

After each half-life, the emission rate is halved. Therefore, we now calculate the emission rate after 3 half-lives:

[tex]\[ \text{Emission rate after 1 half-life} = \frac{4 \times 10^8}{2} = 2 \times 10^8 \][/tex]
[tex]\[ \text{Emission rate after 2 half-lives} = \frac{2 \times 10^8}{2} = 1 \times 10^8 \][/tex]
[tex]\[ \text{Emission rate after 3 half-lives} = \frac{1 \times 10^8}{2} = 0.5 \times 10^8 = 5 \times 10^7 \][/tex]

Thus, the emission rate after 24 days is:
[tex]\[ 5 \times 10^7 \text{ rays per second} \][/tex]

### Summary of Answers
1. (a) The nucleon number of the new nucleus [tex]\(X_e\)[/tex] formed is:
[tex]\[ \boxed{131} \][/tex]

2. (c) The gamma ray emission rate after 24 days is:
[tex]\[ \boxed{5 \times 10^7 \text{ rays per second}} \][/tex]