{"id":2076,"date":"2020-10-25T22:36:36","date_gmt":"2020-10-25T17:36:36","guid":{"rendered":"https:\/\/murreeroad.org\/physics10\/?p=2076"},"modified":"2020-10-26T00:02:49","modified_gmt":"2020-10-25T19:02:49","slug":"physics-10th-chapter-18-numerical-problems","status":"publish","type":"post","link":"https:\/\/murreeroad.org\/physics10\/physics-10th-chapter-18-numerical-problems\/","title":{"rendered":"Physics 10th chapter 18 Numerical Problems"},"content":{"rendered":"

<\/a>Chapter 18<\/strong><\/span><\/p>\n

Atomic and Nuclear Physics <\/strong>(Numerical Problems)<\/strong><\/span><\/h2>\n

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.<\/a><\/strong>
\n18.2. 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?\u00a0\u00a0<\/a><\/strong>
\n18.3. 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?<\/a><\/strong>
\n18.4. 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.\u00a0\u00a0<\/a><\/strong>
\n18.5. 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.<\/a><\/strong><\/p>\n

18.6. In an experiment to measure the half-life of a radioactive element, the following results were obtained:<\/a><\/strong><\/p>\n\n\n\n\n
Count rate \/ minute<\/strong><\/td>\n400<\/strong><\/td>\n200<\/strong><\/td>\n100<\/strong><\/td>\n50<\/strong><\/td>\n25<\/strong><\/td>\n<\/tr>\n
Time (in minutes)<\/strong><\/td>\n0<\/strong><\/td>\n2<\/strong><\/td>\n4<\/strong><\/td>\n6<\/strong><\/td>\n8<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Plot a graph between the count rate and time in minutes. Measure the value for the half-life of the element from the graph.\u00a0\u00a0<\/strong><\/a>
\n18.7. 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.<\/a><\/strong>
\n18.8. 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?<\/a><\/strong><\/p>\n

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?<\/a><\/a><\/strong><\/p>\n

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.<\/strong><\/p>\n

Solution:<\/strong> Let half life of \"\" <\/strong>is T1\/2<\/sub> = 7.3 seconds<\/p>\n

The original activity = A0<\/sub><\/p>\n

As after time T1\/2<\/sub> its activity will become A0<\/sub>\/2<\/p>\n

After 2 T1\/2<\/sub> = 2 x 7.3 = 14.6 seconds. The activity will become A0<\/sub>\/4<\/p>\n

After 3 T1\/2<\/sub> = 3 x 7.3 = 21.9 seconds. The activity will become A0<\/sub>\/8<\/p>\n

After 4 T1\/2<\/sub> = 4 x 7.3 = 28.2 seconds. The activity will become A0<\/sub>\/16<\/p>\n

Hence 1\/16th<\/sup> of the original sample will be left.<\/a><\/p>\n

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?\u00a0\u00a0<\/strong><\/p>\n

Solution: <\/strong>Let half life of Cobalt -60 is T1\/2<\/sub> = 5.25 years<\/p>\n

The original activity = A0<\/sub><\/p>\n

As after time T1\/2 <\/sub>= 5.25 years. The activity will become A0<\/sub>\/2<\/p>\n

After time 2 T1\/2 <\/sub>= 2 x 5.25 = 10.50 years. The activity will become A0<\/sub>\/4<\/p>\n

After time 3 T1\/2 <\/sub>= 3 x 5.25 = 15.75 years. The activity will become A0<\/sub>\/8<\/p>\n

After time 4 T1\/2 <\/sub>= 4 x 5.25 = 21.0 years. The activity will become A0<\/sub>\/16<\/p>\n

After time 5 T1\/2 <\/sub>= 5 x 5.25 = 26.25 years. The activity will become A0<\/sub>\/32<\/p>\n

Hence after 26 years less than 1\/32th<\/sup> of the original sample will be left.<\/a><\/p>\n

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?<\/strong><\/p>\n

Solution: <\/strong>Let half life of Cobalt -14 is T1\/2<\/sub> = 5730 years<\/p>\n

As after time T1\/2 <\/sub>= 5730 years. The activity will become A0<\/sub>\/2<\/p>\n

After time 2 T1\/2 <\/sub>= 2 x 5730 = 11460 years. The activity will become A0<\/sub>\/4<\/p>\n

After time 3 T1\/2 <\/sub>= 3 x 5730 = 17190 years. The activity will become A0<\/sub>\/8<\/p>\n

Hence\u00a0 1\/8th<\/sup> of the original sample will be left after the 1.72 x 104<\/sup> years<\/a><\/p>\n

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.\u00a0\u00a0<\/strong><\/p>\n

Solution: Half life of Technetium = 6 hours<\/strong><\/p>\n

Mass (m0 =200 mg<\/p>\n

How much left after time = 36 hours.<\/p>\n

Mass of Technetium left after first half life T1\/2<\/sub> 6 hours is 100 mg<\/p>\n

Mass of Technetium left after 2nd<\/sup> half life 2 x T1\/2<\/sub> 12 hours is 50 mg<\/p>\n

Mass of Technetium left after 3rd<\/sup> half life 3 x T1\/2<\/sub> 18 hours is 25 mg<\/p>\n

Mass of Technetium left after 4th<\/sup> half life 4 x T1\/2<\/sub> 24 hours is 12.5 mg<\/p>\n

Mass of Technetium left after 5th<\/sup> half life 5 x T1\/2<\/sub> 30 hours is 6.25 mg<\/p>\n

Mass of Technetium left after 6th<\/sup> half life 6 x T1\/2<\/sub> 36 hours is 3.12 mg<\/a><\/p>\n

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.<\/strong><\/p>\n

Solution: <\/strong>Let half life of a radioactive element is T1\/2 <\/sub>= 10mins<\/p>\n

The original activity = A0<\/sub> = 368 counts per minute.<\/p>\n

As after first half time\u00a0 T1\/2 <\/sub>= 10 mins<\/p>\n

The activity will become A0<\/sub>\/2 = 368\/2 = 184 counts per minute<\/p>\n

As after second half life time 2 x T1\/2<\/sub> = 2 x 10 = 20 mins<\/p>\n

The activity will become A0<\/sub>\/4 = 368\/4 = 92 counts per minute<\/p>\n

As after third half life time 3 x T1\/2<\/sub> = 3 x 10 = 30 mins<\/p>\n

The activity will become A0<\/sub>\/8 = 368\/8 = 46 counts per minute<\/p>\n

As after fourth half life time 4 x T1\/2<\/sub> = 4 x 10 = 40 mins<\/p>\n

The activity will become A0<\/sub>\/16 = 368\/16 = 23 counts per minute<\/p>\n

Hence after 40 minutes its count rate will be 23 counts per minute.<\/a><\/p>\n

18.6. In an experiment to measure the half-life of a radioactive element, the following results were obtained:<\/strong><\/p>\n\n\n\n\n
Count rate \/ minute<\/strong><\/td>\n400<\/strong><\/td>\n200<\/strong><\/td>\n100<\/strong><\/td>\n50<\/strong><\/td>\n25<\/strong><\/td>\n<\/tr>\n
Time (in minutes)<\/strong><\/td>\n0<\/strong><\/td>\n2<\/strong><\/td>\n4<\/strong><\/td>\n6<\/strong><\/td>\n8<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Plot a graph between the count rate and time in minutes. Measure the value for the half-life of the element from the graph.\u00a0\u00a0<\/strong><\/p>\n

Ans:<\/strong><\/p>\n

\"\"<\/p>\n

It half life T1\/2<\/sub> is 2 minutes<\/a><\/strong><\/p>\n

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.<\/strong><\/p>\n

Solution: <\/strong>Given data:<\/p>\n

Let half life of a radioactive element is T1\/2 <\/sub>= 1500 years<\/p>\n

The original activity = A0<\/sub> = 3200 counts per minute.<\/p>\n

First half life\u00a0 T1\/2 <\/sub>= 1500 years, activity will be A0<\/sub>\/2 = 3200\/2 = 166 c\/m<\/p>\n

Second half life 2 x\u00a0 T1\/2 <\/sub>= 2 x 1500 = 3000 years, activity will be A0<\/sub>\/4 = 3200\/4 = 800 c\/m<\/p>\n

Third\u00a0 half life 3 x\u00a0 T1\/2 <\/sub>= 3 x 1500 = 4500 years, activity will be A0<\/sub>\/8 = 3200\/8 = 400 c\/m<\/p>\n

Fourth half life 4 x\u00a0 T1\/2 <\/sub>= 4 x 1500 = 6000 years, activity will be A0<\/sub>\/16 = 3200\/16 = 200 c\/m<\/p>\n\n\n\n\n
Count rate\u00a0<\/strong><\/td>\n3200<\/strong><\/td>\n1600<\/strong><\/td>\n800<\/strong><\/td>\n400<\/strong><\/td>\n200<\/strong><\/td>\n<\/tr>\n
Time (in years)<\/strong><\/td>\n0<\/strong><\/td>\n1500<\/strong><\/td>\n3000<\/strong><\/td>\n4500<\/strong><\/td>\n6000<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"\"<\/a><\/p>\n

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?<\/a><\/strong><\/p>\n

Ans: <\/strong>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.<\/p>\n\n\n\n\n
Count rate<\/strong><\/td>\n270<\/strong><\/td>\n280<\/strong><\/td>\n300<\/strong><\/td>\n310<\/strong><\/td>\n285<\/strong><\/td>\n290<\/strong><\/td>\n305<\/strong><\/td>\n315<\/strong><\/td>\n<\/tr>\n
Time succ. 8 hours<\/strong><\/td>\n1<\/strong><\/td>\n2<\/strong><\/td>\n3<\/strong><\/td>\n4<\/strong><\/td>\n5<\/strong><\/td>\n6<\/strong><\/td>\n7<\/strong><\/td>\n8<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

 <\/p>\n

<\/a>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?<\/strong><\/p>\n

Solution: <\/strong>Given data:<\/p>\n

We know half life of C-14 is T1\/2 <\/sub>= 5730 years<\/p>\n

Since the ratio has been reduced by 1\/8 (one-eighth)<\/p>\n

Therefore three half life have passed.<\/p>\n

The age of the fossil is given by first half life\u00a0 T1\/2 <\/sub>= 5730 years<\/p>\n

Age of the fossil is given by after 2nd<\/sup> half life\u00a0 2 x T1\/2 <\/sub>= 2 x 5730 = 14160 years<\/p>\n

Age of the fossil is given by after 3rd<\/sup> half life\u00a0 3 x T1\/2 <\/sub>= 3 x 5730 = 17190 years<\/p>\n

 <\/p>\n

<—————————–><\/p>\n

 <\/p>\n

 <\/p>\n

Goto Top<\/a><\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"

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… Read More »Physics 10th chapter 18 Numerical Problems<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"neve_meta_sidebar":"","neve_meta_container":"","neve_meta_enable_content_width":"off","neve_meta_content_width":0,"neve_meta_title_alignment":"","neve_meta_author_avatar":"","neve_post_elements_order":"","neve_meta_disable_header":"","neve_meta_disable_footer":"","neve_meta_disable_title":""},"categories":[9],"tags":[],"_links":{"self":[{"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/posts\/2076"}],"collection":[{"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/comments?post=2076"}],"version-history":[{"count":5,"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/posts\/2076\/revisions"}],"predecessor-version":[{"id":2084,"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/posts\/2076\/revisions\/2084"}],"wp:attachment":[{"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/media?parent=2076"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/categories?post=2076"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/tags?post=2076"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}