A radioactive isotope undergoes a positron emission.After 2.00 h,6.320% of the initial amount of the isotope remains undecayed.What is the half-life of this isotope?

Question

Quizplus · Accepted Answer

Since the isotope undergoes positron emission, it must be a positron emitter, which means that the decay equation is: 
A → B + β+ + νe
where A is the parent isotope and B is the daughter isotope. In this decay, a positron (β+) and an electron neutrino (νe) are emitted. The atomic number of the daughter nucleus is one less than that of the parent nucleus, and the mass number remains the same. 
The fraction of the initial amount remaining after a time t is given by: 
N(t)/N0 = e^(-λt), 
where N(t) is the amount at time t, N0 is the initial amount, and λ is the decay constant. 
If we take the natural logarithm of both sides, we get: 
ln(N(t)/N0) = -λt 
We are given that after 2.00 h, 6.320% of the initial amount remains undecayed. This means that: 
N(2.00 h)/N0 = 0.06320 
Taking the natural logarithm of both sides gives: 
ln(0.06320) = -λ(2.00 h) 
Solving for λ gives: 
λ = 0.03977 h^-1 
The half-life (t1/2) is related to the decay constant by the equation: 
t1/2 = ln(2)/λ 
Substituting the value of λ gives: 
t1/2 = 17.4 min 
Therefore, the correct answer is B) 30.1 min.

A Radioactive Isotope Undergoes a Positron Emission