etisalat 4G 9:55 PM adpoly.ankabut.ac.ae IC Jasim Alhaddad: Attempt 1 Quiz Note:
ID: 1884741 • Letter: E
Question
etisalat 4G 9:55 PM adpoly.ankabut.ac.ae IC Jasim Alhaddad: Attempt 1 Quiz Note: It is recommended that you save your response as you complete each question. Question 1 (2 points) Manganese-54 has a half life of 312.3 days. If its activity is currently 5430 dpm, what will its activity be 100 days from now? Question 2 (2 points) Using the information in question 1 above, what will be the activity 7 half-lives from now? Question 3 (3 points) A 90Sr source with a half life of 28.78 years was calibrated to emit 0.1 uCi of radiation. If its activity today is measured at 8x 104 dpm, how long was it calibrated? Question 4 (3 points) A 210Po source with a half life of 138.4 days was calibrated at o.1 pCi, 3 years ago. Under your detector it records 2.16 x 103 cpm. What is your detector's percent efficiency?Explanation / Answer
Solution :-
a) Find the activity be 100 days from now,
By using the formula for activity ,
T =( 2.302t/2 )/0.693 * log A0 /A------------(1)
Where , A0 = is the activity at t = 0
A = Activity
t/2 = half life
from equation (1)
0.0963 = log 100 – log A
0.0963-2 = - log A
Log A = 1.9037
A = 80.1124 dpm
The activity be 100 days from now = 80.1124 dpm
b) Find the activity
By using formula ,
t = 7 * t/2
7 * 312.3
t = 2186.1 days
t =( 2.303 *t/2) / 0.693 * log A0 / A
2186.1 = (2.303 * 312.3)/0.693 * log 100 / A
2.1077 = log 100 – log A
0.1077 = -log A
Log A = -0.1077
A = 7.8031 * 10-3 dpm
c) Find how long was it calibrated
t/2 = 28.78 yrs
0.1miceocurie = 2.22*1021*10-6
= 222*1015
A = 8 * 104 dpm
t = ?
t = (2.303 t/2)/0.693 * log A0/A
t = ( 2.303 * 28.78 ) / 0.693 * log 2.22*1015/8*104
t = 998.8210 years
d) Find detector percentage efficiency
t/2 = 138.4 days
t= 3 years
% efficiency = No. of counts rcorded by detector / No. of radiation coming out from source .
= Nc / Ns
= 2.16*103 cpm / 2.22*1015cpm
% efficiency = 9.7297 * 10-13%
Detector percentage efficiency is = 9.7297 * 10-13%
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