Chapter 18

Atomic and Nuclear Physics (Numerical Problems)

18.1. The half-life of is is 7.3 s. A sample of this nuclide of nitrogen is observed for 29.2 s. Calculate the fraction of the original radioactive isotope remaining after this time.
18.2. Cobalt-60 is a radioactive element with half-life of 5.25 years. What fraction of the original sample will be left after 26 years?  
18.3. Carbon-14 has a half-life of 5730 years. How long will it take for the quantity of carbon-14 in a sample to drop to one-eighth of the initial quantity?
18.4. Technetium-99 m is a radioactive element and is used to diagnose brain, thyroid, liver and kidney diseases. This element has half-life of 6 hours. If there is 200 mg of this technetium present, how much will be left in 36 hours.  
18.5. Half-life of a radioactive element is 10 minutes. If the initial count rate is 368 counts per minute, find the time for which count rates reaches 23 counts per minute.

18.6. In an experiment to measure the half-life of a radioactive element, the following results were obtained:

Count rate / minute 400 200 100 50 25
Time (in minutes) 0 2 4 6 8

Plot a graph between the count rate and time in minutes. Measure the value for the half-life of the element from the graph.  
18.7. A sample of certain radioactive element has a half-life of 1500 years. If it has an activity of 32000 counts per hour at the present time, then plot a graph of the activity of this sample over the period in which it will reduce to 1/16 of its present value.
18.8. Half-life of a radioactive element was found to be 4000 years. The count rates per minute for 8 successive hours were found to be 270, 280, 300, 310, 285, 290, 305, 312. What does the variation in count rates show? Plot a graph between the count rates and time in hours. Why the graph is a straight line rather than an exponential?

18.9. Ashes from a campfire deep in a cave show carbon-14 activity of only one-eighth the activity of fresh wood. How long ago was that campfire made?

18.1. The half-life of is is 7.3 seconds. A sample of this nuclide of nitrogen is observed for 29.2 s. Calculate the fraction of the original radioactive isotope remaining after this time.

Solution: Let half life of is T1/2 = 7.3 seconds

The original activity = A0

As after time T1/2 its activity will become A0/2

After 2 T1/2 = 2 x 7.3 = 14.6 seconds. The activity will become A0/4

After 3 T1/2 = 3 x 7.3 = 21.9 seconds. The activity will become A0/8

After 4 T1/2 = 4 x 7.3 = 28.2 seconds. The activity will become A0/16

Hence 1/16th of the original sample will be left.

18.2. Cobalt-60 is a radioactive element with half-life of 5.25 years. What fraction of the original sample will be left after 26 years?  

Solution: Let half life of Cobalt -60 is T1/2 = 5.25 years

The original activity = A0

As after time T1/2 = 5.25 years. The activity will become A0/2

After time 2 T1/2 = 2 x 5.25 = 10.50 years. The activity will become A0/4

After time 3 T1/2 = 3 x 5.25 = 15.75 years. The activity will become A0/8

After time 4 T1/2 = 4 x 5.25 = 21.0 years. The activity will become A0/16

After time 5 T1/2 = 5 x 5.25 = 26.25 years. The activity will become A0/32

Hence after 26 years less than 1/32th of the original sample will be left.

18.3. Carbon-14 has a half-life of 5730 years. How long will it take for the quantity of carbon-14 in a sample to drop to one-eighth of the initial quantity?

Solution: Let half life of Cobalt -14 is T1/2 = 5730 years

As after time T1/2 = 5730 years. The activity will become A0/2

After time 2 T1/2 = 2 x 5730 = 11460 years. The activity will become A0/4

After time 3 T1/2 = 3 x 5730 = 17190 years. The activity will become A0/8

Hence  1/8th of the original sample will be left after the 1.72 x 104 years

18.4. Technetium-99 m is a radioactive element and is used to diagnose brain, thyroid, liver and kidney diseases. This element has half-life of 6 hours. If there is 200 mg of this technetium present, how much will be left in 36 hours.  

Solution: Half life of Technetium = 6 hours

Mass (m0 =200 mg

How much left after time = 36 hours.

Mass of Technetium left after first half life T1/2 6 hours is 100 mg

Mass of Technetium left after 2nd half life 2 x T1/2 12 hours is 50 mg

Mass of Technetium left after 3rd half life 3 x T1/2 18 hours is 25 mg

Mass of Technetium left after 4th half life 4 x T1/2 24 hours is 12.5 mg

Mass of Technetium left after 5th half life 5 x T1/2 30 hours is 6.25 mg

Mass of Technetium left after 6th half life 6 x T1/2 36 hours is 3.12 mg

18.5. Half-life of a radioactive element is 10 minutes. If the initial count rate is 368 counts per minute, find the time for which count rates reaches 23 counts per minute.

Solution: Let half life of a radioactive element is T1/2 = 10mins

The original activity = A0 = 368 counts per minute.

As after first half time  T1/2 = 10 mins

The activity will become A0/2 = 368/2 = 184 counts per minute

As after second half life time 2 x T1/2 = 2 x 10 = 20 mins

The activity will become A0/4 = 368/4 = 92 counts per minute

As after third half life time 3 x T1/2 = 3 x 10 = 30 mins

The activity will become A0/8 = 368/8 = 46 counts per minute

As after fourth half life time 4 x T1/2 = 4 x 10 = 40 mins

The activity will become A0/16 = 368/16 = 23 counts per minute

Hence after 40 minutes its count rate will be 23 counts per minute.

18.6. In an experiment to measure the half-life of a radioactive element, the following results were obtained:

Count rate / minute 400 200 100 50 25
Time (in minutes) 0 2 4 6 8

Plot a graph between the count rate and time in minutes. Measure the value for the half-life of the element from the graph.  

Ans:

It half life T1/2 is 2 minutes

18.7. A sample of certain radioactive element has a half-life of 1500 years. If it has an activity of 32000 counts per hour at the present time, then plot a graph of the activity of this sample over the period in which it will reduce to 1/16 of its present value.

Solution: Given data:

Let half life of a radioactive element is T1/2 = 1500 years

The original activity = A0 = 3200 counts per minute.

First half life  T1/2 = 1500 years, activity will be A0/2 = 3200/2 = 166 c/m

Second half life 2 x  T1/2 = 2 x 1500 = 3000 years, activity will be A0/4 = 3200/4 = 800 c/m

Third  half life 3 x  T1/2 = 3 x 1500 = 4500 years, activity will be A0/8 = 3200/8 = 400 c/m

Fourth half life 4 x  T1/2 = 4 x 1500 = 6000 years, activity will be A0/16 = 3200/16 = 200 c/m

Count rate  3200 1600 800 400 200
Time (in years) 0 1500 3000 4500 6000

18.8. Half-life of a radioactive element was found to be 4000 years. The count rates per minute for 8 successive hours were found to be 270, 280, 300, 310, 285, 290, 305, 312. What does the variation in count rates show? Plot a graph between the count rates and time in hours. Why the graph is a straight line rather than an exponential?

Ans: Variation in count rate shows the random nature of radioactive decay, graph is almost horizontal line rather than exponential curve which is due to long half-life as compared to period of 8 hours.

Count rate 270 280 300 310 285 290 305 315
Time succ. 8 hours 1 2 3 4 5 6 7 8

 

18.9. Ashes from a campfire deep in a cave show carbon-14 activity of only one-eighth the activity of fresh wood. How long ago was that campfire made?

Solution: Given data:

We know half life of C-14 is T1/2 = 5730 years

Since the ratio has been reduced by 1/8 (one-eighth)

Therefore three half life have passed.

The age of the fossil is given by first half life  T1/2 = 5730 years

Age of the fossil is given by after 2nd half life  2 x T1/2 = 2 x 5730 = 14160 years

Age of the fossil is given by after 3rd half life  3 x T1/2 = 3 x 5730 = 17190 years

 

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