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Two radioactive isotopes P and Q have half-lives 10 minutes and 15 minutes respectively. Freshly prepared samples of each isotope initially contain the same number of atoms. After 30 minutes, the ratio \(\rm \dfrac{number \ of \ atoms \ of \ P}{number \ of \ atoms \ of \ Q}\) will be
1. 0.5
2. 2.0
3. 1.0
4. 3.0

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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 N: 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

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