# During World War II, tritium ( 3 H) was a component of fluorescent watch dials and hands. Assume you have such a watch that was made in January 1944. If 17% or more of the original tritium was needed to read the dial in dark places, until what year could you read the time at night? (For 3 H, t 1/2 = 12.3 yr.)

### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

Chapter
Section

### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 18, Problem 65AE
Textbook Problem
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## During World War II, tritium (3H) was a component of fluorescent watch dials and hands. Assume you have such a watch that was made in January 1944. If 17% or more of the original tritium was needed to read the dial in dark places, until what year could you read the time at night? (For 3H, t1/2 = 12.3 yr.)

Interpretation Introduction

Interpretation: Use of tritium as a component of fluorescent watch dials and its half life is given. The year upto which the time at night can be read using the given kind of watch 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. Tritium was used as a component of fluorescent watch dials during world war 2

To determine: The year up to which the time at night can be read using the given kind of watch.

### Explanation of Solution

Explanation

The decay constant is calculated by the formula given below.

λ=0.693t1/2

Where

• t1/2 is the half life of nuclide.
• λ is decay constant.

The value of t1/2 is 12.3years .

Substitute the value of half life in the above formula.

λ=0.69312.3year-1=0.05634years-1_ .

The decay constant is 0.05634year-1_

Explanation

Decay time is calculated by the formula,

t=2.303λlog(n0n)

Where

• n0 is the number of atoms initially present.
• n is the number of atoms remaining after decay.
• t is the decay time.

Fraction of tritium remaining is 17n0100 .

Therefore, n=17n0100

Substitute the values of n , n0 and decay constant in the above equation.

t=2

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