The half life of 92 U 238 is 4-5 -109 years- 3 g of active atoms were found after 18 -109 years- Then the amount of active atoms originally present was-

The correct option is D 48 g Half life = 4.5 ×10^9 years,Time interval = 18 ×10^9 years ∴ #160;Number of half life periods n = 18 ...

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Standard XII

Chemistry

Molecular Speeds - Vrms, Vavg, Vmps

The half life...

Question

The half life of $_{92} {U^{238}}$ is $4.5 \times 10^{9}$ years. $3 g$ of active atoms were found after $18 \times 10^{9}$ years. Then the amount of active atoms originally present was:

4 \times 10^{9} g

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12 g

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24 g

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48 g

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Solution

The correct option is D $48 g$
Half life $= 4.5 \times 10^{9}$ years,
Time interval $= 18 \times 10^{9}$ years
$∴$ Number of half life periods $n = \frac{18 \times 10^{9}}{4.5 \times 10^{9}} = 4$
Now, $N = N_{0} {(\frac{1}{2})}^{n}$
$3 = N_{0} {(\frac{1}{2})}^{4} = \frac{N_{0}}{16}$
$∴ N_{0} = 16 \times 3 = 48 g m s$

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The half life of 92U238 is 4.5×109 years. 3g of active atoms were found after 18×109 years. Then the amount of active atoms originally present was:

The correct option is D 48gHalf life =4.5×109 years,Time interval =18×109 years∴ Number of half life periods n=18×1094.5×109=4Now, N=N0(12)n 3=N0(12)4=N016∴N0=16×3=48gms

The half life of $_{92} {U^{238}}$ is $4.5 \times 10^{9}$ years. $3 g$ of active atoms were found after $18 \times 10^{9}$ years. Then the amount of active atoms originally present was: