Deduce the expression, N = N₀e⁻λt, for the law of radioactive decay. Write symbolically the process expressing the β⁺ decay of ²²₁₁Na. Also write the basic nuclear process underlying this decay. Is the nucleus formed in the decay of the nucleus ²²₁₁Na, an isotope or isobar?

Let N₀ = Total number of atoms present originally in a sample at time t = 0
N = Total number of atoms left undecayed in the sample at time t
dN = A small number of atoms that disintegrate in a small interval of time dt
Rate of disintegration of the element, R = -dN/dt = λN
λ = disintegration constant
dN/N = -λdt
Integrate both side:
∫dN/N = ∫-λdt
logeN = -λt + C
C = constant of integration
t = 0, N = N₀
logeN = -λ × 0 + C ⇒ C = logeN₀
logeN = -λt + logeN₀
loge(N/N₀) = -λt
N/N₀ = e⁻λt
N = N₀e⁻λt
(B) (i). The basic nuclear process underlying this β⁺ decay.
p → n + e⁺ + v
The reaction is:
²²₁₁Na → β⁺ + ²²₁₀Ne + v
(ii) It is an isobar (same mass number but different atomic number).