Correct Answer - Option 1 : 0.5
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
\(⇒ \frac{N}{N_{0}} = (\frac{1}{2})^{n}\)
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 - \(t_{\frac{1}{2}}\) = TP = 10 minute, \(t_{\frac{1}{2}}\)= TQ = 15 minute
After 30 minutes the number of half-lives is given by
For P, nP = \(\frac{30}{10} = 3\) ,
For Q, nQ = \(\dfrac{30}{15} = 2\)
- The corresponding number of particles left for P and Q are
\(⇒ N_P = N_0 \left(\dfrac{1}{2}\right)^3 \ \ \ \ \ \ N_Q = N_0 \left(\dfrac{1}{2}\right)^2\)
The ration NP : NQ is given by
\(⇒ \frac{N_{P}}{N_{Q}} = \frac{N_{0}\times (\frac{1}{2})^{3}}{N_{0}\times (\frac{1}{2})^{2}} = \frac{4}{8} = \frac{1}{2}\)
⇒ NP ∶ NQ = 1 ∶ 2
\(⇒ \dfrac{N_P}{N_Q}=0.5\)
- Hence, option 1 is the answer