Derive the relationship between half life and decay constant of a radi

1.	Derive the relationship between half life and decay constant of a radioelement.
Answer» Equation for the decay constant is given as, λ = \(\frac{2.303}{t}\) log₁₀\(\frac{N_0}{N}\) …(i) Where, λ = Decay constant N = Number of nuclei (atoms) present at time t At t = 0, N = N₀. Hence, At t = t_1/2, N = N₀/2 Substitution of these values of N and t in equation (i) gives, λ = \(\frac{2.303}{t_{1/2}}\) log₁₀\(\frac{N_0}{\frac{N_0}{2}}\) = \(\frac{2.303}{t_{1/2}}\)log₁₀² = \(\frac{2.303}{t_{1/2}}\) x 0.3010 = \(\frac{0.693}{t_{1/2}}\) Hence, λ = \(\frac{0.693}{t_{1/2}}\) or t_1/2 =\(\frac{0.693}{\lambda}\)

Derive the relationship between half life and decay constant of a radioelement.

Answer»

Equation for the decay constant is given as,

λ = \(\frac{2.303}{t}\) log₁₀\(\frac{N_0}{N}\) …(i)

Where,

λ = Decay constant

N = Number of nuclei (atoms) present at time t

At t = 0,

N = N₀.

Hence,

At t = t_1/2,

N = N₀/2

Substitution of these values of N and t in equation (i) gives,

λ = \(\frac{2.303}{t_{1/2}}\) log₁₀\(\frac{N_0}{\frac{N_0}{2}}\)

= \(\frac{2.303}{t_{1/2}}\)log₁₀²

= \(\frac{2.303}{t_{1/2}}\) x 0.3010

= \(\frac{0.693}{t_{1/2}}\)

Hence,

λ = \(\frac{0.693}{t_{1/2}}\) or t_1/2 =\(\frac{0.693}{\lambda}\)