Answered

Discover the answers you need at Westonci.ca, where experts provide clear and concise information on various topics. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

It took 55 days for a radioactivity of 1.75 x 1012 Bq to remain 0.135 Ci. What is the half-life of this radioactivity?

Sagot :

pstnonsonjoku

From the calculations, we can see that the half life of the material is 11.4 days

What is radioactivity?

The term radioactivity refers to the spontaneous disintegration of a radioactive nucleus. The time taken for half of the radioactive material to remain.

Given that;

N/No = (1/2)^t/t1/2

N = amount that remained

No = amount that remained in the beginning

t = time taken

t1/2 = half life of the substance

N =Noe^-kt

4.995 ×10^9 = 1.75 x 10^12e^-55k

3.5* 10-2=e^-55k

k = ln3.5* 10-2/-55

k =0.06095

t1/2 = 0.693/0.06095

t1/2 = 11.4 days

Learn more about half life:https://brainly.com/question/24710827

#SPJ1

We hope our answers were useful. Return anytime for more information and answers to any other questions you have. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.