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Physics Part-II

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The normal activity of living carbon-containing matter is foun

Physics Part-II

Class 12

NCERT

Chapter 1: Ray Optics and Optical Instruments

41 questions

Chapter 2: Wave Optics

30 questions

Chapter 3: Dual Nature of Radiation and Matter

44 questions

Chapter 4: Atoms

23 questions

Chapter 5: Nuclei

38 questions

A radioactive isotope has a half-life of

T

years. How long will it take the activity to reduce to (a)

3.125%

(b)

1%

of its original value?

The normal activity of living carbon-containing matter is found to be about 15 decays per ute for every gram of carbon. This activity arises from the small proportion of radioactive

_{6}^{14} C

present with the stable carbon isotope

_{6}^{12} C

. When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life ( 5730 years) of

_{6}^{14} C

, and the measured activity, the age of the specimen can be approximately estimated. This is the principle of

_{6}^{14} C

dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per ute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.

Obtain the amount of

_{27}^{60} C o

necessary to provide a radioactive source of 8.0 mCi strength. The half-life of

_{27}^{60} C o

is 5.3 years.

View all question

Chapter 6: Semiconductor Electronics: Materials, Devices and Simple Circuits

25 questions

Question

Medium

Solving time: 5 mins

The normal activity of living carbon-containing matter is found to be about 15 decays per ute for every gram of carbon. This activity arises from the small proportion of radioactive $_{6}^{14} C$ present with the stable carbon isotope $_{6}^{12} C$ . When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life ( 5730 years) of $_{6}^{14} C$ , and the measured activity, the age of the specimen can be approximately estimated. This is the principle of $_{6}^{14} C$ dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per ute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.

Uploaded: 2022-05-11T16:58:00.000Z
Duration: 5 min

Views: 6,150 students

Updated on: May 11, 2022

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Physics Part-II (NCERT)

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The normal activity of living carbon-containing matter is found to be about 15 decays per ute for every gram of carbon. This activity arises from the small proportion of radioactive

_{6}^{14} C

present with the stable carbon isotope

_{6}^{12} C

. When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life ( 5730 years) of

_{6}^{14} C

, and the measured activity, the age of the specimen can be approximately estimated. This is the principle of

_{6}^{14} C

dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per ute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.

8 mins ago

Discuss this question LIVE

8 mins ago

Text solutionVerified

Suppose that the number of

C - 14

atoms per gram where

N_{0}

t = 0

(initially) when its activity was

15

decays per minute per gram. Today ie., after time

t

(Age of once living organism), the number of C-14 atom per gram left is

N

and shows an activity of 9 decays per minute per gram. Now, activity of a radioactive substance is given by

R = R_{0} e^{- λ t}

R_{0} = 15 d ec a ys m^{- 1}; R = 9 d ec a ys m^{- 1}

∴ 9 = 15 \times e^{- λ t} \Rightarrow e^{- λ t} = \frac{15}{9} = 1.6667

\Rightarrow λ t = lo g_{e} 1.6667 = 2.303 lo g_{10} 1.6667 = 2.303 \times 0.2219 = 0.5110 \Rightarrow t = \frac{0.5110}{λ}

Now decay constant

λ = \frac{0.693}{T} = \frac{0.693}{5 , 730} ye a r^{- 1}

∴ t = \frac{0.5110 \times 5730}{0.693} = 4225.15 ye a rs

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Practice more questions from Physics Part-II (NCERT)

A radioactive isotope has a half-life of

T

years. How long will it take the activity to reduce to (a)

3.125%

(b)

1%

of its original value?

The normal activity of living carbon-containing matter is found to be about 15 decays per ute for every gram of carbon. This activity arises from the small proportion of radioactive

_{6}^{14} C

present with the stable carbon isotope

_{6}^{12} C

. When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life ( 5730 years) of

_{6}^{14} C

, and the measured activity, the age of the specimen can be approximately estimated. This is the principle of

_{6}^{14} C

dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per ute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.

Obtain the amount of

_{27}^{60} C o

necessary to provide a radioactive source of 8.0 mCi strength. The half-life of

_{27}^{60} C o

is 5.3 years.

View all

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Question Text	The normal activity of living carbon-containing matter is found to be about 15 decays per ute for every gram of carbon. This activity arises from the small proportion of radioactive $_{6}^{14} C$ present with the stable carbon isotope $_{6}^{12} C$ . When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life ( 5730 years) of $_{6}^{14} C$ , and the measured activity, the age of the specimen can be approximately estimated. This is the principle of $_{6}^{14} C$ dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per ute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.
Updated On	May 11, 2022
Topic	Nuclei
Subject	Physics
Class	Class 12
Answer Type	Text solution:1 Video solution: 2
Upvotes	257
Avg. Video Duration	7 min

Text solutionVerified

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