Question

Half-lives for α and β emission of radioactive material are 16 years and 48 years respectively. When material decays giving α and β emission simultaneously then time in which \(\frac{3}{4}^{th}\) of the material decays is
1. 29 years
2. 24 years
3. 64 years
4. 12 years

Answer 1

Correct Answer - Option 2 : 24 years

CONCEPT: 

• Half-life is the time required for the nucleus to decay half of the initial concentration
• If the same nuclide is simultaneously decaying into two nuclides in half-lives T₁ and T₂ then the effective half-life of the decay is given by

\(\Rightarrow T =\frac{T_1T_2}{T_1+T_2}\)

Where T₁ = First half-life, T₂ = Second half-life

• If N0 is the initial number of atoms of nuclide present after n number of half-lives the amount of substance left is given by

\(⇒ \frac{N}{N_{0}} = (\frac{1}{2})^{n}\)

Where N = The number of atoms left after n number of half-lives, N0 = The initial number of atoms,n = The number of half-lives  [ n = \(\frac{T}{t_{\frac{1}{2}}}\) Where T = Time up to which the disintegration observed and \(t_{\frac{1}{2}}\) =Half life]

CALCULATION :

Let, T₁ = 16 years, T₂ = 48 Years 

• Effective half-life

\(⇒ T =\frac{T_1T_2}{T_1+T_2}= \frac{16× 48}{16+48} = 12 years\)

• The amount of substance left after  \(\frac{3}{4}^{th}\) of the decay is  

⇒ N = N₀ - \(\frac{3}{4}\)

⇒ N = 1  - \(\frac{3}{4}\) = \(\frac{1}{4}\)

• The number of half-lives required for the decay is given by

\(⇒ \frac{N}{N_0}=\left(\frac{1}{2}\right)^n\)

\(\Rightarrow \frac{1}{4}=\left(\frac{1}{2}\right)^n\)

\(\Rightarrow (\frac{1}{2})^{2} = (\frac{1}{2})^{n}\)

Comparing both the equations n = 2

The time required for the decay will be 

⇒ T^; = n× T    [ We know T = \(t_\frac{1}{2}\)]

Hence, the time required for the decay will be 

 T^,  = 2× T = 2× 12 = 24 years

Hence, time will be 24 years

• Hence, option 2 is the answer