SATHEE: Nuclei - Result Question 65

Nuclei - Result Question 65

66. A mixture consists of two radioactive materials $A_{1}$ and $A_{2}$ with half lives of $20 s$ and $10 s$ respectively. Initially the mixture has $40 g$ of $A_{1}$ and $160 g$ of $A_{2}$ . The amount of the two in the mixture will become equal after :

(a) $60 s$

(b) $80 s$

(d) $40 s$

[2012]

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Answer:

Correct Answer: 66. (d)

Solution:

(d) Let, the amount of the two in the mixture will become equal after $t$ years.

The amount of $A_{1}$ , which remains after $t$ years

$N_{1} = \frac{N_{01}}{(2)^{t / 20}}$

The amount of $A_{2}$ , which remains, after $t$ years

$N_{2} = \frac{N_{02}}{(2)^{t / 10}}$

According to the problem

$N_{1} = N_{2}$

$\frac{40}{(2)^{t / 20}} = \frac{160}{(2)^{t / 10}}$

$2^{t / 20} = 2^{(\frac{t}{10} - 2)}$

$\frac{t}{20} = \frac{t}{10} - 2$

$\frac{t}{20} - \frac{t}{10} = 2$

$\frac{t}{20} = 2$

$t = 40 s$

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Nuclei - Result Question 65

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

Answer:

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66. A mixture consists of two radioactive materials $A_{1}$ and $A_{2}$ with half lives of $20 s$ and $10 s$ respectively. Initially the mixture has $40 g$ of $A_{1}$ and $160 g$ of $A_{2}$ . The amount of the two in the mixture will become equal after :