00:01
In this question we need to solve for decay constant.
00:03
So the decay constant lambda can be found by using half -life formula which is given by lambda as a reference to ln2 lnhir represent natural log and t half.
00:15
Half -life is already given to us.
00:17
Half -life of 14 carbon is 5 ,730 years.
00:22
First we need to convert this time in eov.
00:26
T -half is equal to 57730.
00:30
So, we'll be converting it into days, first of all, and then it will be converting to hours, then minutes, and then now finally in second, okay? this one is for days, then hour, then minute, and then second.
00:43
So whatever you answer will be, that will be in second.
00:46
So we'll be getting approximate answer as 1 .808 into 10 to be power 11 seconds.
00:55
So now we have to calculate decay constant.
00:57
So, lambda is equal to ln2 upon 1 .808 into 10 to 3 power 11 seconds.
01:07
So, for converting this term to 0 .303, then log at the base 10 and 2, and the no meter will remain same.
01:15
This will be 2 .303.
01:18
This value can be log 10, can be capped, you can see through, log 2 value in natural ways you can see on log table.
01:27
And you will get the value so that is 0 .3010.
01:31
Again when you solve this you'll be getting an approximate answer as 3 .834 into 10 to be power minus 12 second energies.
01:39
Okay so this is a part answer now we'll be calculating similarly for b part...