   Chapter 19, Problem 33E

Chapter
Section
Textbook Problem

# Technetium-99 has been used as a radiographic agent in bone scans ( T 43 99 c is absorbed by bones). If T 43 99 c has a half-life of 6.0 hours, what fraction of an administered dose of 100. μg T 43 99 c remains in a patient’s body after 2.0 days?

Interpretation Introduction

Interpretation: Fraction of administered dose of 100μg4399Tc remaining in someone’s body after 2.0 days is to be calculated.

Concept introduction: A process through which an unstable nuclide looses its energy due to excess of protons or neutrons is known as radioactive decay. The cause of instability of a nuclide is its inefficiency in holding the nucleus together. The curie is commonly used to measure nuclear radioactivity.

4399Tc is absorbed by the bones and is used as a radiographic agent in bone scans.

To determine: The fraction of administered dose of 100μg4399Tc remaining in someone’s body after 2.0 days

Explanation

Explanation

The given time is 2days.

Therefore, the time in hours =24h1day=48h

Half-life of technetium is 6h.

The fraction 4399Tc is calculated by the formula given below.

ln[NN0]=Ktln[NN0]=[0.693t1/2]×t

Where

• N0 is the final amount of technetium.
• N is the final amount of technetium.
• t1/2 is the half-life of technetium.
• t is the time of decay.

Substitute the values in the above equation.

ln[NN0]=[0

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