Radioactiv edecay will occur as follows `overset(220)86Rnrarroverset(216)84PO + overset(4)2He Half life =55s` `overset(216)84Porarroverset(212)82Pb +
Radioactiv edecay will occur as follows
`overset(220)86Rnrarroverset(216)84PO + overset(4)2He Half life =55s`
`overset(216)84Porarroverset(212)82Pb + overset(4)2He Half life =0.66s`
`overset(812)82Pbrarroverset(212)82BL + gamma^(@)(4)e Half life =10.6 h`
If a certain mass of radon (Rn=220) is allowed to decay in a certain container,then after 5 minutes the element with the greater mass will be
A. radon
B. polonium
C. lead
D. bismuth
1 Answers
Correct Answer - C
Let `M_(0)` be intial mass of Rn-220 Number of half lives of Rn in 5 munuites `=(5min)/(55s)=(5xx60)/(55)=5.5`
`therefore Mass of Rn-220(left)=(1)/(2)^(5.5) M_(0)=(M_(0)/(45)`
`therefore The mass converted in to Po-216 is `(44//45)`
`M_(0)` Number of hlf lives of Po-216 in 5 minutes
`=(5min)/(0.66)=(300)/(0.66)=455`
`therefore Mass of Po-216left=((1)/(2))^(455)xx(44)/(45)M_(0)rarr0`
This means that Po-216 formed will soon decay to lead Hence lead will have maximum mass nearly `(44)/(45)M_(0)`