A human body required 0.01 m activity of radioactive substance after 24 hours .Half li of radioactive substance is 6 hours .Then injection of maximum activity of radioactive substance that can be injected is :
a). 0.08
b). 0.04
c). 0.16
d). 0.32
six hours is four halflives.
.1=(1/2)^4 * original=original/16
solve for original
To find the maximum activity of the radioactive substance that can be injected, we need to understand the concept of half-life.
Half-life is the time it takes for half of a substance to decay. In this case, the half-life of the radioactive substance is given as 6 hours.
Let's work through the problem step by step:
1. Since the half-life is 6 hours, after 6 hours, the activity of the substance in the body will be reduced to half of its original value.
Initial activity: 0.01 m
After 6 hours: 0.01 m / 2 = 0.005 m
2. After another 6 hours (total of 12 hours), the activity will be reduced to half of the value after 6 hours.
Previous activity: 0.005 m
After 12 hours: 0.005 m / 2 = 0.0025 m (or 2.5 mBq)
3. After another 6 hours (total of 18 hours), the activity will again be halved.
Previous activity: 0.0025 m
After 18 hours: 0.0025 m / 2 = 0.00125 m (or 1.25 mBq)
4. Finally, after a total of 24 hours (the given time period), the activity will again be halved.
Previous activity: 0.00125 m
After 24 hours: 0.00125 m / 2 = 0.000625 m (or 0.625 mBq)
Therefore, after 24 hours, the activity of the radioactive substance in the human body will be 0.000625 m (or 0.625 mBq).
The maximum activity of the radioactive substance that can be injected is the activity that will result in a final value of 0.01 m after 24 hours.
So, we need to find the injection activity that will result in an activity of 0.01 m after 24 hours. Let's call this value "x".
Starting activity (injected activity): x
After 24 hours: x / 2 / 2 / 2 / 2 = x / 16
We need x / 16 to be equal to 0.01 m.
x / 16 = 0.01 m
Solving for x:
x = 16 * 0.01 m
x = 0.16 m (or 160 mBq)
Therefore, the maximum activity of the radioactive substance that can be injected is 0.16 m (or 160 mBq).
Thus, the correct answer is (c) 0.16.