NEET Class neet Answered
please answer this
Asked by Prashant DIGHE | 11 Jan, 2020, 09:45: PM
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
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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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