. Atomic mass of


83Bi209 is 208.980388 u and that of 26Fe56 is 55.934939 u. Which nucleus is more stable? Given that mass of a proton= 1.007825 u and mass of a neutron = 1.008665 u

A. Both are equally stable 

B. 83Bi209 is more stable 

C. Both are unstable nuclei 

D. 26Fe56 is more stable 

Best Answer

Ans: D

Sol: First we have to calculate binding energy per nucleon of both nuclei. 

(a) For 83Bi209 :- 
Δm = (126 mn + 83mp) - m 83Bi209 
= (127.09179+83.649475) – 208.980388 
= 1.760877 
Binding Energy = Δm ×931=1639.37 MeV 
B.E per nucleon = 1639.37/209=7.84 MeV 
(b) For 26Fe56 :- 
Δm = (30 mn + 26 mn) – m 26Fe56 
= (30.25995+26.20345) – 55.934939 
= 0.528461 
Binding energy= Δm×931=491.99 MeV 
B.E per nucleon = 491.99/56=8.78 MeV 
B.E per nucleon of 26Fe56 is more than that of 83Bi209 
Therefore, 26Fe56 is more stable than 83Bi209

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