Cesium 124 has a half-life of 31. what fraction of cesium 124 sample will remain after 0.10h
Answers
Answer:
We have to find the friction of siege in 1-4 sample will remain after point an hour and the half is given solution Given dirty half is equal to 31 seconds And that time strategical to 0.10 hours. So this is converting in 0.10 is multiplied by 3600. This is converting in second so have you know that 0.693 a phone T half this is the value of constant lambda is equal to 2.303 upon T log off and not upon Anti by solving this. G 0.693 The half is totally 1/2. It is called to 2.303 Upon 0.10 into 3600 seconds log off and not upon And the and now by solving this we get the values that This is 0.693 multiplied by 0.10 is multiplied by 3600 upon 31 in multiply 2.303 is equal to log off and not upon angry and by solving this we get log off and not upon. Nt it's equal to this is 3.49446. And we can write this in form of and not upon. NT It's because to 10 days to cover 3.49 446 points to 3090 29 five points so by solving this. Nt is exposed to And not upon 309 0.295 is a question of zero Jiro Jiro Jiro burrito. And so the answer should be there. It is Anti upon and not supposed to 0.0003. So the answer is should be, the main value is the 0.0003.