Correct Answer - Option 2 : 12 h
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 - t1/2 = 2 hours, N0 = 64 [To work safely the number of reacting molecules must decrease by 64 times]
- The amount of substance left after n number of half-lives is
\(\Rightarrow N=N_0 \left(\dfrac{1}{2}\right)^n\)
The above equation can be rewritten as
\(⇒ \dfrac{N}{N_0} = \left(\dfrac{1}{2}\right)^n\)
\(⇒ \dfrac{1}{64}=\left(\dfrac{1}{2}\right)^n\)
\(⇒ (\frac{1}{2})^{6} = (\frac{1}{2})^{n}\)
Comparing both sides of the equation n = 6
- The time after which it is safe to work with the source will be
⇒ Time = n× \(t_{\frac{1}{2}}\)
Substituting the given values in the above equation
⇒ Time = 6 × 2 = 12 hours
- The minimum time after which it would be possible to work safely with this source is 12 hours
- Hence, option 2 is the answer