A mixture consists of two radioactive materials A1 andA2 with half lives of 20 s and 10 s respectively. Initially the mixture has 40 g of A1 and 160 g of A2. The amount of the two in the mixture will become equal after:

  • Option 1)

    60 s

  • Option 2)

    80 s

  • Option 3)

    20 s

  • Option 4)

    40 s

 

Answers (1)

For A1, N01=40g, t1/2= 20 sec

For A2, N02=160g, t1/2 = 10 sec

Let us say at t =t, amount of A and Abecome equal then at t = t

N_{1}=\frac{N_{01}}{2\frac{t}{20}}    &      N_{2}=\frac{N_{02}}{2\frac{t}{10}}

\because N_{1}= N_{2} \ and \ N_{01}=40, N_{02}=160

\Rightarrow \frac{40}{2\frac{t}{20}}=\frac{160}{2\frac{t}{10}} \Rightarrow 4 =2^{\frac{t}{10}-\frac{t}{20}}=2^{\frac{t}{20}}

or \frac{t}{20}=2 \ or \ t=40 sec


Option 1)

60 s

This option is incorrect.

Option 2)

80 s

This option is incorrect

Option 3)

20 s

Option 4)

40 s

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