Radioactive decay will occur as follows aligned & ₈₆²²⁰ Rn ₈₄²¹⁶ PO + ₂^4 He Half life =55 ~s & ₈₄⁸¹⁶ Po ₈₂²¹² ~Pb + ₂^4 He Half life =0 66 ~s & ₈₂²¹² BL + ^ e Half life =10 6 ~h aligned If a certain mass of radon (R n=220) is allowed to decay in a certain container; then after 5 minutes the element with the greater mass will be

Step-by-step solution

Let M₀ be initial mass of R n-220 aligned & Number of half-lives of R n in 5 minutes = 5 ~min 55 ~s & = 5 60 55 5.5 aligned Mass of R n-220 (left) = ( 1 2 )^ 5.5 M₀= M₀ 45 The mass converted into Po -216 is (44/45) M₀ , Number of half-lives of P o-216 in 5 minutes = 5 ~min 0.66 = 300 0.66 455 Mass of P o-216 left = ( 1 2 )⁴⁵⁵ 44 45 M₀ 0 This means that P o-216 formed will soon decay to lead. Hence, lead will have maximum mass nearly ( 44 45 ) M₀