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Question 2 12 pts A radioactive isotope of cesium, 137Cs, which is common in spent nuclear fuel and in nuclear contamination, can you solve and show work
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This question is solved by using simple concept of radioactive decay which follows first order kinetics.

The answer is given in the image,

Answer: CAL = 39.88g. ter z 27-58 years K = 0.693 > 0.693 tha 27.58 k=0-02513 year I t = 177.8 years k= 2.303 log CALO 0.025139.88 = 87. Lo CAIT CAJ+ = 39.88 87.10 CA]t = 0.45799. 0.4579 g of 1316 is left.

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