Solved Numerical Problem No. 21.2 ( Unit-21:Nuclear Physics); HSSC-II (Class-12); Physics (Punjab Curriculum and Text Board, Lahore) With Verified Answers

The half-life of’ `\ _{38}^{91}Sr` is 9.70 hours. Find its decay constant. (Ans: 1.99 ｘ 10⁻⁵ s)

Given Data:

half life of radioactive nucleus `\ _{86}^{226}Ra` = `\T_{1∕2}` = 9.70 hours

by converting into Sec (SI unit) we have 1 hour = 3.6 ｘ 10³ s  

`\T_{1∕2}` = 9.70 ｘ 3.6 ｘ 10³ s  = 3.492 ｘ 10³ s = 3.492 ｘ 10⁴ s

To Find:

The decay constant = λ =?

Solution:

Using the formula for Half-life we have

`\T_{1∕2}` = `\frac {0.693}{λ}`

or

 λ = `\frac {0.693}{T_{1∕2}}`

Putting values

 λ = `\frac {0.693}{3.492 ｘ 10⁴ s}`

 λ = 0.1985 ｘ 10⁻⁴ s⁻¹

or

 λ = 1.99 ｘ 10⁻⁵ s⁻¹ ----------------Ans.

Similar Questions:

The half life of radioactive nucleus `\ _{86}^{226}Ra` is 1.6 ｘ10³ years. Determine the decay constant. (Answer: 1.4 ｘ 10⁻¹¹ s⁻¹)

************************************

Shortcut Links For 

1. National MDCAT, ECAT, JOB EXAM Physics MCQs Preparation (New) 
2. Physics (New) 11th Class SLO Based Notes
3. Physics (New) 11th class MCQs Quizzes All Units
4. Physics (New) 12th Class SLO Based Notes
5. Physics (New) 12th class MCQs Quizzes All Units

1. Website for School and College Level Physics   

(https://askrarequiz.blogspot.com)

2. Website for School and College Level Mathematics  

(https://askraremaths.blogspot.com)

3. Website for Single National Curriculum Pakistan - All Subjects Notes 

(https://askrareclassnotes.blogspot.com/)

© 2022-Onwards by Academic Skills and Knowledge (ASK)     

Note:  Write me in the comment box below for any query and also Share this information with your class-fellows and friends.