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natta225 [31]
3 years ago
7

As part of her nuclear stress test, Simone starts to walk on the treadmill. When she reaches the maximum level, Paul injects a r

adioactive dye with201Tl with an activity of 72 MBq . The radiation emitted from areas of her heart is detected by a scanner and produces images of her heart muscle. After Simone rests for 3 h, Paul injects more dye with 201Tl and she is placed under the scanner again. A second set of images of her heart muscle at rest is taken. When Simone's doctor reviews her scans, he assures her that she had normal blood flow to her heart muscle, both at rest and under stress.
Part A
If the half-life of 201Tl is 3.0 days, what is its activity, in megabecquerels after 3.0 days?
Express the activity to two significant figures and include the appropriate units.

Part B
What is its activity in megabecquerels after 6.0 days?
Express the activity to two significant figures and include the appropriate units.
Chemistry
1 answer:
Naya [18.7K]3 years ago
3 0

Answer:

Part A: 36 MBq; Part B: 18 MBq

Explanation:

The half-life is the time it takes for half the substance to disappear.

The activity decreases by half every half-life

A =Ao(½)^n, where n is the number of half-lives.

Part A

3.0 da = 1 half-life

A =  Ao(½) = ½ × 72 MBq = 36 MBq

Part B

6.0 da = 2 half-lives

A = Ao(½)^2 = ¼ × 72 MBq = 18 MBq

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Explanation:

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So, the rate constant k value is:

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Substitute the given values and the k value in this formula to get the concentration of the reactant after time 8 hrs is shown below:

\frac{0.693}{1.5 hrs} =\frac{2.303}{8 hrs} x log \frac{8.4x10^-^5}{y} \\ log \frac{8.4x10^-^5}{y} =1.604\\\frac{8.4x10^-^5}{y}=10^1^.^6^0^4\\\frac{8.4x10^-^5}{y}=40.18\\y=\frac{8.4x10^-^5}{40.18} \\=>y=2.09x10^-^6

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