NEET Class neet Answered
A→→C(Half life=1 minute), B→→C(Half life=2minutes). At t=0, concentration of A was 4N? whereas concentration of B was N? at t=0. Find- (i) the concentration of C at the instance when concentration of A and B becomes equal. Also find(ii) the concentration of C at the instance when activity of A becomes equal to activity of B.
Asked by patra04011965 | 30 Mar, 2020, 20:02: PM
Concentration of A at time t ,
........................ (1)
![begin mathsize 14px style N subscript A equals space 4 space N subscript o space e to the power of negative lambda subscript A t end exponent space end style](https://images.topperlearning.com/topper/tinymce/cache/750a315afaa122ddf1966bb833f072ec.png)
Concentration of B at time t ,
.........................( 2 )
![begin mathsize 14px style N subscript B equals space space N subscript o space e to the power of negative lambda subscript B space t end exponent space end style](https://images.topperlearning.com/topper/tinymce/cache/8911e4ddcba8aec2b39d2ccbffc4c270.png)
where λA is decay constant of A and it is given by, λA = ln2 min-1
where λB is decay constant of B and it is given by, λB = (1/2) ln2 min-1
Time t for getting equal concentration of A and B is obtained by equating above equation and solving for time t
If NA = NB , then we have , ![begin mathsize 14px style 4 space N subscript o space e to the power of negative lambda subscript A t end exponent space space equals space N subscript o space e to the power of negative lambda subscript B t end exponent space end style](https://images.topperlearning.com/topper/tinymce/cache/43e73ef24a9b0e87f6c776972af62655.png)
![begin mathsize 14px style 4 space N subscript o space e to the power of negative lambda subscript A t end exponent space space equals space N subscript o space e to the power of negative lambda subscript B t end exponent space end style](https://images.topperlearning.com/topper/tinymce/cache/43e73ef24a9b0e87f6c776972af62655.png)
![begin mathsize 14px style ln left parenthesis 4 right parenthesis space minus space lambda subscript A t space equals space minus lambda subscript B t end style](https://images.topperlearning.com/topper/tinymce/cache/6987b5b1d2db45fbd17cbb6cc96a9480.png)
![begin mathsize 14px style t space equals space fraction numerator ln left parenthesis 4 right parenthesis over denominator lambda subscript A minus lambda subscript B end fraction space equals space fraction numerator ln left parenthesis 4 right parenthesis over denominator ln left parenthesis 2 right parenthesis space minus space begin display style 1 half end style ln left parenthesis 2 right parenthesis end fraction space equals space 4 space m i n end style](https://images.topperlearning.com/topper/tinymce/cache/6b058ab5c71e3dd18ac7db1097b3c5a4.png)
Hence concentration of C at t = 4 min given by,
![begin mathsize 14px style C subscript t equals 4 end subscript space equals space 4 N subscript o space e to the power of negative space 4 space ln 2 end exponent space plus space N subscript o space e to the power of negative 2 space ln 2 end exponent space equals N subscript o subscript space end subscript open square brackets 4 over 16 space plus space 1 fourth close square brackets space equals space N subscript o over 2 end style](https://images.topperlearning.com/topper/tinymce/cache/87fbbc15f33a2e3e386a9fd8e5784759.png)
----------------------------------------------------------
Activity of A , ![begin mathsize 14px style fraction numerator d space N subscript A over denominator d t end fraction space equals space minus lambda subscript A N subscript A equals space minus lambda subscript A space 4 space N subscript o space e to the power of negative lambda subscript A t end exponent end style](https://images.topperlearning.com/topper/tinymce/cache/eaa65a3cb14982ac9ebf7d5bcf1e2538.png)
![begin mathsize 14px style fraction numerator d space N subscript A over denominator d t end fraction space equals space minus lambda subscript A N subscript A equals space minus lambda subscript A space 4 space N subscript o space e to the power of negative lambda subscript A t end exponent end style](https://images.topperlearning.com/topper/tinymce/cache/eaa65a3cb14982ac9ebf7d5bcf1e2538.png)
Activity of B , ![begin mathsize 14px style fraction numerator d space N subscript B over denominator d t end fraction space equals space minus lambda subscript B N subscript B equals space minus lambda subscript B space N subscript o space e to the power of negative lambda subscript B t end exponent end style](https://images.topperlearning.com/topper/tinymce/cache/9cb12fc3e07cbbb5b06c5971d7765303.png)
![begin mathsize 14px style fraction numerator d space N subscript B over denominator d t end fraction space equals space minus lambda subscript B N subscript B equals space minus lambda subscript B space N subscript o space e to the power of negative lambda subscript B t end exponent end style](https://images.topperlearning.com/topper/tinymce/cache/9cb12fc3e07cbbb5b06c5971d7765303.png)
If both activities are equal, then we have,
![begin mathsize 14px style lambda subscript A space 4 space N subscript o space e to the power of negative lambda subscript A t end exponent space equals space lambda subscript B space N subscript o space e to the power of negative lambda subscript B t end exponent space end style](https://images.topperlearning.com/topper/tinymce/cache/c4737fd17132d400ec1d8db3d3be39e8.png)
![begin mathsize 14px style e to the power of open parentheses lambda subscript A space minus space lambda subscript B close parentheses t end exponent space equals space 4 space lambda subscript A over lambda subscript B space equals space 8 end style](https://images.topperlearning.com/topper/tinymce/cache/c09f593f120161fcd867e9951745ae51.png)
![begin mathsize 14px style open parentheses lambda subscript A space minus space lambda subscript B close parentheses space t space equals space ln left parenthesis 8 right parenthesis space space space o r space space t space equals space fraction numerator ln left parenthesis 8 right parenthesis over denominator open parentheses lambda subscript A space minus space lambda subscript B close parentheses end fraction space equals space fraction numerator 2 space ln left parenthesis 8 right parenthesis over denominator ln left parenthesis 2 right parenthesis end fraction space equals space 6 space m i n end style](https://images.topperlearning.com/topper/tinymce/cache/d41126e5e3d5d515813f14d6340969e7.png)
Hence concentration of C at t = 6 min given by,
![begin mathsize 14px style C subscript t equals 6 end subscript space equals space 4 N subscript o space e to the power of negative space 6 space ln 2 end exponent space plus space N subscript o space e to the power of negative 3 space ln 2 end exponent space equals N subscript o subscript space end subscript open square brackets 4 over 64 space plus space 1 over 8 close square brackets space equals space fraction numerator 3 N subscript o over denominator 16 end fraction end style](https://images.topperlearning.com/topper/tinymce/cache/7e9fa3050c4cede101f16b2118dfe6ef.png)
Answered by Thiyagarajan K | 30 Mar, 2020, 22:40: PM
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