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Physics Part-II

Nuclei

Obtain the amount of ^{60}_{27}Co necessary to provide a radi

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

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.

The half-life of

_{38}^{90} S r

is 28 years. What is the disintegration rate of 15 mg of this isotope?

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

25 questions

Question

Easy

Solving time: 4 mins

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.

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Updated on: Apr 7, 2023

Book:

Physics Part-II (NCERT)

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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.

8 mins ago

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8 mins ago

Text solutionVerified

Given,

T_{1/2} = 5.3 ye a r

T_{1/2} = 5.3 \times 365 \times 24 \times 60 \times 60 = 1.67 \times 1 0^{8} sec

\frac{d N}{d t} = 8.0 m C I

\frac{d N}{d t} = 8 \times 1 0^{- 3} \times 3.7 \times 1 0^{10} = 29.6 \times 1 0^{7} d ec a y / sec

Half- life,

T_{1/2} = \frac{0.693}{λ}

λ = \frac{0.693}{T _{1/2}} = \frac{0.693}{1.67 \times 1 0 ^{8}}

λ = 0.414 \times 1 0^{- 8} s^{- 1}

Radioactive rate,

\frac{d N}{d t} = λ N

N = \frac{1}{λ} . \frac{d N}{d t}

N = \frac{29.6 \times 1 0 ^{7}}{0.414 \times 1 0 ^{- 8}} = 7.15 \times 1 0^{16} a t o m

For

_{27}^{60} C o

mass of

6.022 \times 1 0^{23} a t o m = 60 g

mass of one atom

= \frac{60}{6.022 \times 1 0 ^{23}} g

mass of one atom

7.15 \times 1 0^{16} a t o m \equiv \frac{60 \times 7.15 \times 1 0 ^{16}}{6.022 \times 1 0 ^{23}} g = 7.123 \times 1 0^{- 6} g

Hence, the amount of

_{27}^{60} C o

7.123 \times 1 0^{- 6} g

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Practice more questions from Physics Part-II (NCERT)

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.

The half-life of

_{38}^{90} S r

is 28 years. What is the disintegration rate of 15 mg of this isotope?

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Question Text	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.
Updated On	Apr 7, 2023
Topic	Nuclei
Subject	Physics
Class	Class 12
Answer Type	Text solution:1 Video solution: 6
Upvotes	699
Avg. Video Duration	7 min

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.

Text solutionVerified

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