00:01
Hello, in the question we have given that a radioactive isotope has an activity of 8 .83 into 10 raised to 4 bacquerel initially.
00:11
So, the initial activity, so that is a0 is equal to 8 .83 into 10 raised to 4 bacquerel is given.
00:25
So, after 3 .35 hours, so that is time t is equal to 3 .35 hours, the activity is 5 .39.
00:40
So, a is equal to 5 .39 into 10 raised to 4 bacquerel.
00:51
So, what is the lifetime of the isotope? so, in the first part, we have to find out the lifetime or half -life.
01:08
Now, in order to find this, we will use the equation of activity.
01:13
So, we know that the activity is given by a is equal to a0 e raised to minus lambda t.
01:27
So, now let us plug the value, a is given to be 5 .39 into 10 raised to 4 that is equal to 8 .83 into 10 raised to 4 into e raised to minus this decay constant we don't know.
01:50
So, it is minus lambda times t is 3 .35.
01:55
So, this 10 raised to 4 will go away and 5 .39 divided by 8 .83 will turn out to be 0 .61.
02:06
So, this e raised to minus lambda times 3 .35 will be equal to 0 .61.
02:18
So, taking log on both the sides, so taking natural log that is ln on both sides, what we get? so, we get it as minus lambda times 3 .35 is equals to ln of 0 .61.
02:45
So, from here, we will get this as minus of lambda times 3 .35 is equals to minus of 0 .49.
02:57
So, this minus minus will go away.
02:59
So, lambda will be equal to 0 .1462 and the unit will be our inverse.
03:09
So, this is the decay constant and once we know the decay constant, we can solve many things.
03:16
So, we know that the time period t half is 0 .693 divided by lambda...