A fresh radioactive mixture has short lived species A and B- Both emit - - particles initially at 8000 particles per minute- 20 minutes later- they emit 3500 - - particles per minute- If the half - lives of the species A and B are 10 minutes and 500 hours respectively- then the ratio of activities of A - B in the initial mixture was-A- 4- 6B- 6- 4 - 3- 4D- 3- 1

The correct option is D 3 : 1 nbsp; nbsp; nbsp; nbsp; nbsp; A→ X + ^4_2He t=0 nbsp; nbsp; nbsp;A_o t=20 nbsp; nbsp;A_o/4 nbsp; ...

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Standard XI

Mathematics

Definition of Relations

A fresh radio...

Question

A fresh radioactive mixture has short lived species $A$ and $B$ . Both emit $α$ − particles initially at 8000 particles per minute. 20 minutes later, they emit 3500 $α$ − particles per minute. If the half – lives of the species $A$ and $B$ are 10 minutes and 500 hours respectively, then the ratio of activities of A : B in the initial mixture was:

4 : 6

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6 : 4

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3 : 4

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3 : 1

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Solution

The correct option is D $3 : 1$
$A \to X +_{2}^{4} H e$
t=0 $A_{o}$
t=20 $\frac{A_{o}}{4}$

$B \to Y +_{2}^{4} H e$
t=0 $B_{o}$
t=20 $B_{o}$
As, $t_{\frac{1}{2}}$ for $B$ is very high its amount can be assumed to remain remain same after 20 minutes.

$A_{o} + B_{o}$ = 8000
$\frac{A_{o}}{4} + B_{o}$ = 3500
$A_{o}$ = 6000
$B_{o}$ = 2000
$\frac{A_{o}}{B_{o}}$ = $\frac{3}{1}$

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The correct option is D 3:1 A→ X + 42He t=0 Ao t=20 Ao4 B→ Y + 42He t=0 Bo t=20 Bo As, t12 for B is very high its amount can be assumed to remain remain same after 20 minutes. Ao+Bo = 8000 Ao4+Bo = 3500 Ao = 6000 Bo = 2000 AoBo = 31