0 Christina Perri-Pengu + × -2 Calendar x Y G Indium-192 Is One Radioi xy-Imperi

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0 Christina Perri-Pengu + × -2 Calendar x Y G Indium-192 Is One Radioi xy-Imperial Valley College- -) Dwww.saplinglearning.com/ib.scms/mod/ibis/view.php?id=1914984 Jump to... Logout O Help sapling learning Sapling Learning My Assignment Imperial Valley College-CHEM 100-Fall 15-BEN EDICTO Activities and Due Dates HW 17 Resources 12/4/2015 11:55 PM A 15.1/20 Assignment Information Available From 11/20/2015 12:00 AM Due Date: Points Possible: Grade Category: Graded Description: Policies: Gradebook #AttemptsScore Print CalculatorPeriodic Table Question 11 of 17 12/4/2015 11:55 PM 20 sapling learning The half-lives of different radioisotopes are given in the table below. How long would it take (in minutes) for the amount of lead-196 to decrease from 76.0 mg to 9.50 mg? Homework Radioisotope Half-life (min) argon-44 lead-196 potassium-44 indium-117 You can check your answers. You can view solutions when you complete or give up on any question. 37 You can keep trying to answer each question until you get it right or give up. 43 You lose 5% of the points available to each answer in your question for each incorrect attempt at that answer Number min Help With This Topic Web Help &Videos; Technical Support and Bug Reports 12 13 15 Previous & Give Up & View Solution O Check Answer Next Exit Hint Copyright © 2011-2015 Sapling Learning, Inc.-204 about us careers partners privacy policy terms of use contact ushelp

Explanation / Answer

we know that

decay constant (lamda) = 0.693 / ( t1/2)

given

t1/2 for lead = 37

so

lamda = 0.693 / 37

lamda = 0.018734

now

we know that

for a radioactive reaction

C = Co x e^(-lamda x t)

apply logarithm on both sides

ln C = ln Co - (lamda x t)

lamda x t = ln Co - ln C

lamda x t = ln ( Co / C)

given

Co = 76

C = 9.5

so

0.018734 x t = ln ( 76 / 9.5)

t = 111

so

the time taken is 111 minutes

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