Question

The decay constant λ, the half life \(T_{\frac{1}{2}}\) and the mean life τ of a radioactive nucleus are related as:  
1. \(\tau = \dfrac{1}{\lambda}\ ; \ T_{\frac{1}{2}} = 0.693 \ \lambda\)
2. \(\tau = \dfrac{1}{\lambda} \ ; \ T_{\frac{1}{2}}=\tau . \ln 2\)
3. \(\tau = \dfrac{\ln 2}{\lambda} \ ; \ T_{\frac{1}{2}} = \tau . \ln 2\)
4. \(\tau = \lambda . \ln 2 \ ; T_{\frac{1}{2}} = \dfrac{\ln 2}{\lambda}\)

Answer 1

Correct Answer - Option 2 : \(\tau = \dfrac{1}{\lambda} \ ; \ T_{\frac{1}{2}}=\tau . \ln 2\)

The correct answer is option 2) i.e. \(τ = \dfrac{1}{λ} \ ; \ T_{\frac{1}{2}}=τ . \ln 2\)

CONCEPT:

• Radioactive decay: It is the process of spontaneous breakdown of the nucleus of a radioactive atom, which results in the emission of radiation from the nucleus.

The formula to represent radioactive decay is:

N = N0e-λT 

Where No is the initial quantity of the substance, N is the remaining quantity that is not yet decayed, T is the elapsed time of decay and λ is the decay rate constant.

CALCULATION:

At half lifetime, the amount of quantity left equals half of the initial quantity. ⇒ N = 0.5 N₀

N = N0e-λT

0.5N₀ = N₀e^-λT ⇒ 0.5 = e-λT ⇒ ln(0.5) = -λT

⇒ -0.693 = -λT

⇒ T = \(\frac{0.693}{λ } = \frac{ln2}{λ}\)

Therefore, the half-life (T1/2) of a substance is the time taken by its nucleus to decay to half of its initial quantity.  \(T_{1/2} = \frac{ln\:2}{λ}\)      ----(1)

Mean life or average life of a radioactive element, τ = total lifetime of all atoms ÷ total number of atoms

τ = \(\frac{\int_0^\infty TdN}{N_0} = \frac{\int_0^\infty Td(N_{0}e^{-λ T} )}{N_0} = \frac{\int_0^\infty -λTN_0e^{-λ T}dT}{N_0}\)

= -λ\(\int_0^\infty Te^{-λ T}dT\) = -λ \([ \frac{Te^{-\lambda T}}{-\lambda}]_0 ^\infty - [\frac{e^{-\lambda T}}{-\lambda} dt]\)

= \(\frac{-1}{\lambda} [e^{-\infty} - e^{-0}] = \frac{-1}{\lambda} [0 - 1] \)

τ = \(\frac{1}{\lambda}\)      ----(2)

Substituting (2) in (1)

 \(T_{1/2} = \tau.{ln\:2}\)