U 92238 is known to undergo radioactive decay to form Pb 82206 by...

{}_{92}^{238}U{\to}_{82}^{206}Pb{+}_{2}^{4}He + 6{}_{- 1}^{{}^{0}}\beta$</p><p>$0 . 286 \times {10}^{- 6}mole$</p><p>\left[0 . 286 \times {10}^{- 6} \times {N}_{a}\right]$

$= 0 . 286 \times 6 . 02 \times {10}^{17}$

$= 1 . 72 \times {10}^{17}$

After 3 half lives

$N = \dfrac{No}{( 2{)}^{3}} = \dfrac{1 . 72 \times {10}^{17}}{8}$

$= 0 . 215 \times {10}^{17}$

So, no. of molecule of uranium decayed

$= [ 1 . 72 - 0 . 215 ] \times {10}^{17}$

$= 1 . 5 \times {10}^{17 -}$

So No. of $\alpha$ particle produced $= 8 \times 1 . 5 \times {10}^{17}$

$= 12 \times {10}^{17}$

$= 1 . 2 \times {10}^{18}$