Nuclei with magic no. of proton Z=2,8,20,28,50,52 and magic no. of neutrons N=2,8,20,28,50,82 and 126 are found to be stable. (i) Verify this by calculating the proton separation energy Sp for. 120Sn ( Z = 50 ) and 121Sb( Z = 51 ). The proton separation energy for a nuclide is the minimum energy required to separated the least tightly bound proton form a nucleus of that nuclide. It is given by Sp = ( Mz-1 , N + MHβˆ’Mz ,N )c2 𝑆𝑝 = ( 𝑀𝑧 - 1 , 𝑁 + 𝑀H – 𝑀Z , 𝑁 ) 𝑐 2 . given 119𝑆𝑛 = 118.9058 𝑒 , . 120𝑆𝑛 = 199.902199Β  , . 121𝑆𝑏 = 120.903824 𝑒 ,1𝐻 = 1.0078252 𝑒 (ii) what does the existence of magic number indicate?

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8 months ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- B.E = (Mp+MH-MN)c2

B.E= (118.9058+1.0078252-119.902199)c2

B.E=0.0114362 c2

B.E= (Mp+MH-MN)c2

B.E= (119.902199+1.0078252-120.902822)c2

B.E= 0.0059912c2

Β  (ii) the existence of magic numbers indicates that the shell structure o

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Physics Ncert Solutions Class 12th 2023

Physics Ncert Solutions Class 12th 2023

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