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Asked by Prashant DIGHE | 11 Jan, 2020, 21:45: PM
answered-by-expert Expert Answer
Before decay mass of parent nucleus = M+Δm
 
After decay net mass of two daughter's nuclei = (M/2) + (M/2) = M
 
Hence energy released in the form of  ( Δm c2 ) . 
 
Hence Binding energy of parent nucleus is greater than sum of Binding energy of daughter's nuclei
 
E1 ( M + Δm ) > E2 [ (M/2) + (M/2) ] 
E1 ( M + Δm ) > E2 M
 
( E1 / E2 ) > M / ( M + Δm )
 
if [ M / ( M + Δm ) ] = α ,  and α < 1
 
Hence ( E1 / E2 ) > α  but ( E1 / E2 ) ≈ 1
 
Above condition is satisfied only if, E1 > E2
-----------------------------------------------------------------------
 
If the released energy ( Δm c2 ) is shared by two daughter's nuclei equally,
then energy of each daughter nuclei = (1/2)( Δm c2 )
 
Hence kinetic energy of each daughter nuclei , 
 
(1/2) (M/2) v2  = (1/2)( Δm c2 ) ........................(1)
 
where v is speed of daughter nucleus.
 
From above eqn.(1), we get  v = c [ 2 (Δm / M) ]1/2
 
 
Answered by Thiyagarajan K | 12 Jan, 2020, 10:08: AM

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