Physics · Nuclear Physics
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A sample of F-18 is used internally as a medical diagnostic tool to look for the effect of the positron decay (T1/2 = 110 min), how long does it take for 99% of the F-18 to decay?
- A
10.2 h
- B
11.2 h
- C
14.2 h
- D
12.2 h
From law of radioactivity, N = Noe^-(lamda)(t)
Here decay constant = 0.693/ t ½ = 0.693/110
Decay constant (lamda) = 0.0063min^-
Let No = 100 so rest amount after decay at t
N = (100-99) = 1
So now 1= 100e^(-0.0063)t
Or it can also be written as
e^0.0063t = 100
0.0063t = ln100 = ln10^2 = 2 ln 10
Thus , t = ln100/ln2 x 100
t = 730 min = 12.16 hrs = 12.2 hours
Mathematically other options are incorrect
this option is incorrect. 10.2 h, 11.2 h, and 14.2 h: These options do not accurately reflect the time required for 99% of F-18 to decay, as they don't align with the exponential decay pattern dictated by the half-life.
this option is incorrect. 10.2 h, 11.2 h, and 14.2 h: These options do not accurately reflect the time required for 99% of F-18 to decay, as they don't align with the exponential decay pattern dictated by the half-life.
this option is incorrect. 10.2 h, 11.2 h, and 14.2 h: These options do not accurately reflect the time required for 99% of F-18 to decay, as they don't align with the exponential decay pattern dictated by the half-life.
From law of radioactivity, N = Noe^-(lamda)(t)
Here decay constant = 0.693/ t ½ = 0.693/110
Decay constant (lamda) = 0.0063min^-
Let No = 100 so rest amount after decay at t
N = (100-99) = 1
So now 1= 100e^(-0.0063)t
Or it can also be written as
e^0.0063t = 100
0.0063t = ln100 = ln10^2 = 2 ln 10
Thus , t = ln100/ln2 x 100
t = 730 min = 12.16 hrs = 12.2 hours
Mathematically other options are incorrect
Tagged under Physics · Nuclear Physics · 2021