A radioactive isotope Z A m t 1 - 2 -10 days decays to give Z-6 B m-12 stable atom along with -particles- If m g of - A- are taken and kept in a sealed tube- how much - He- will accumulate in 20 days at STP-

_ Z A ^ m ⟶ _ Z 6 B ^ m 12 +3[ _ 2 He ^ 4 ] Mole of A= m m =1 Number of half lives = 20 / 10 =2 N= N _ 0 ( 1 2 #160;) #16 ...

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Standard XII

Chemistry

Nuclear Stability

A radioactive...

Question

A radioactive isotope $_{Z} A^{m}$ ( $t_{1 / 2} = 10$ days) decays to give $_{Z - 6} B^{m - 12}$ stable atom along with $α$ -particles. If $m g$ of ' $A$ ' are taken and kept in a sealed tube, how much ' $H e$ ' will accumulate in $20$ days at STP?

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Solution

$_{Z} A^{m} ⟶$ $_{Z - 6} B^{m - 12} + 3 [_{2} {H e}^{4}]$
Mole of $A = \frac{m}{m} = 1$
Number of half lives $= 20 / 10 = 2$
$N = N_{0} {(\frac{1}{2})}^{n} = 1 {(\frac{1}{2})}^{2} = \frac{1}{4}$
Decayed moles $= 1 - 1 / 4 = 3 / 4$
Moles of ' $H e$ ' formed $= 3 \times 3 / 4 = 9 / 4$
Volume of ' $H e$ ' at STP $= 22.4 \times \frac{9}{4} = 50.4$ litre

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A radioactive isotope ZAm (t1/2=10 days) decays to give Z−6Bm−12 stable atom along with α-particles. If mg of 'A' are taken and kept in a sealed tube, how much 'He' will accumulate in 20 days at STP?

ZAm⟶ Z−6Bm−12+3[2He4]Mole of A=mm=1Number of half lives =20/10=2N=N0(12)n=1(12)2=14Decayed moles =1−1/4=3/4Moles of 'He' formed =3×3/4=9/4Volume of 'He' at STP =22.4×94=50.4 litre

A radioactive isotope $_{Z} A^{m}$ ( $t_{1 / 2} = 10$ days) decays to give $_{Z - 6} B^{m - 12}$ stable atom along with $α$ -particles. If $m g$ of ' $A$ ' are taken and kept in a sealed tube, how much ' $H e$ ' will accumulate in $20$ days at STP?