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Physics Part-II

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The half-life of ^{90}_{38}Sr is 28 years. What is the disinte

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

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?

Obtain approximately the ratio of the nuclear radii of the gold isotope

_{79}^{197} A u

and the silver isotope

_{47}^{107} A g

View all question

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

25 questions

Question

Medium

Solving time: 5 mins

The half-life of $_{38}^{90} S r$ is 28 years. What is the disintegration rate of 15 mg of this isotope?

Name: Video solution 1: [Solved] The half-life of 3890Sr is 28 years. What is the disintegration rate of 15 mg of this isotope? - Physics Part-II
Uploaded: 2023-09-06T05:40:47.000Z
Duration: 2 min
Description: Video solution 1: [Solved] The half-life of 3890Sr is 28 years. What is the disintegration rate of 15 mg of this isotope? - Physics Part-II

Views: 6,252 students

Updated on: Sep 6, 2023

Book:

Physics Part-II (NCERT)

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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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Text solutionVerified

Given,

m = 15 m g

T_{1/2} = 28 ye a rs = 28 \times 365 \times 24 \times 60 \times 60 = 8.83 \times 1 0^{8} sec

Half-life ,

T_{1/2} = \frac{0.693}{λ}

λ = \frac{0.693}{T _{1/2}}

λ = \frac{0.693}{8.83 \times 1 0 ^{8}} = 7.84 \times 1 0^{- 10}

15 m g

_{38}^{90} S r

contaion number of atoms

N = \frac{15 \times 1 0 ^{- 3} \times 6.022 \times 1 0 ^{23}}{90} = 1.003 \times 1 0^{20} a t o m

Rate of integration,

\frac{d N}{d t} = λ N

\frac{d N}{d t} = 7.84 \times 1 0^{- 10} \times 1.003 \times 1 0^{20}

\frac{d N}{d t} = 7.87 \times 1 0^{10} a t o m s / s

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Practice more questions from Physics Part-II (NCERT)

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?

Obtain approximately the ratio of the nuclear radii of the gold isotope

_{79}^{197} A u

and the silver isotope

_{47}^{107} A g

View all

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Question Text	The half-life of $_{38}^{90} S r$ is 28 years. What is the disintegration rate of 15 mg of this isotope?
Updated On	Sep 6, 2023
Topic	Nuclei
Subject	Physics
Class	Class 12
Answer Type	Text solution:1 Video solution: 2
Upvotes	281
Avg. Video Duration	6 min

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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The half-life of $_{38}^{90} S r$ is 28 years. What is the disintegration rate of 15 mg of this isotope?