Physics · nuclei · NEET
N is the number of nuclei still left (not yet decayed). As time passes this number falls, so its rate of change dN/dt must be negative. The quantity lambda and N are both positive, so we put a minus sign to make the right side negative and match the physical fact that N decreases. If you drop the minus sign you would be saying nuclei are being created, which is wrong.
Separate the variables: dN/N = -lambda dt. Integrate both sides: integral of dN/N gives ln N, and integral of -lambda dt gives -lambda t + C. So ln N = -lambda t + C. At t = 0, N = N0, so C = ln N0. Then ln N - ln N0 = -lambda t, which is ln(N/N0) = -lambda t. Taking exponential of both sides gives N/N0 = e^(-lambda t), so N = N0 e^(-lambda t).
In the pure decay law N = N0 e^(-lambda t), N is the NUMBER of undecayed nuclei. But the same exponential shape works for mass (m = m0 e^(-lambda t)) and for activity (A = A0 e^(-lambda t)), because mass is proportional to number and activity A = lambda N is also proportional to number. So all three decay with the same lambda and same half-life. Just be careful which quantity the question gives you.
No. Radioactive decay comes from inside the nucleus, so lambda is fixed for a given nuclide and does not change with temperature, pressure, or whether the atom is in a compound. This is a favourite NEET point. Only the number of nuclei present affects how many decay per second, not the outside conditions.
e^(-lambda t) is the fraction of the original nuclei that are STILL present after time t. At t = 0 it equals 1 (all present). As t grows it shrinks toward 0. When t = 1/lambda (the mean life), e^(-1) = 0.37, so about 37 percent are left. It never becomes exactly zero, which is why we use half-life to describe the pace instead of a total finish time.
Because the number decaying each second is proportional to how many are left. When many nuclei are present, many decay per second, so N drops steeply. As N gets smaller, fewer decay per second, so the fall slows down. This gives a curve that is steep at first and flattens later - an exponential decay curve, not a straight line.
The rate of decay of a radioactive sample is directly proportional to the number of undecayed nuclei present at that instant: dN/dt = -lambda N, giving N = N0 e^(-lambda t).
Since lambda t must be dimensionless, lambda has units of per second (s^-1). It represents the fraction of nuclei that decay per unit time.
Set N = N0/2 in N = N0 e^(-lambda t). This gives 1/2 = e^(-lambda T), so lambda T = ln 2, meaning half-life T = 0.693/lambda. Half-life and lambda are inversely related.
No. Radioactive decay is a random (statistical) process. The law only predicts the behaviour of a large number of nuclei on average; it cannot say when any one particular nucleus will decay.
Yes. Since activity A = lambda N, multiplying the decay law by lambda gives A = A0 e^(-lambda t). Activity falls with the same decay constant and half-life as the number of nuclei.
