Question

After 3 hours, only 0.25 mg of a pure radioactive material is left. If initial mass was 2 mg then the half life of the substance is
1. 1.5 hr
2. 1 hr
3. 0.5 hr
4. 2 hr

Answer 1

Correct Answer - Option 2 : 1 hr

CONCEPT: 

• 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

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

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 :

Given - initial mass, N₀ = 2 mg, Final mass.N = 0.25 mg, T = 3 hours 

• The number of nuclides left after n number of half-lives is given as 

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

Substituting the given values in the above equation 

\(\Rightarrow\frac{0.25}{2}=\left(\frac{1}{2}\right)^n\)

\(\Rightarrow \frac{1}{8}=\left(\frac{1}{2}\right)^n\)

\(\Rightarrow (\frac{1}{2})^{3} = (\frac{1}{2})^{n}\)

Comparing both the equations, n= 3

The number of half-lives is 3, n= 3

• The relationship between the number of half-lives and time and the half-life is given as  

\(\Rightarrow n=\frac{T}{t_{\frac{1}{2}}}\)

The above equation can be rewritten as

\(\Rightarrow t_{\frac{1}{2}} = \frac{T}{n}\)

Substituting the values 

\(\Rightarrow t_{\frac{1}{2}} = \frac{3}{3} = 1 \ \ Hours\)

• Hence, option 1 is the answer