Ans D Sol First we have to calculate binding energy per nucleon of both nuclei a For 83Bi209 m 126 mn 83mp m 83Bi209 127 09179 83 649475 208 980388 1 760877 Binding Energy m 931 1639 37 MeV B E per nucleon 1639 37 209 7 84 MeV b For 26Fe56 m 30 mn 26 mn m 26Fe56 30 25995 26 20345 55 934939 0 528461 Binding energy m 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

. 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 m_n + 83m_p) - 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 m_n + 26 m_n) – 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

. Atomic mass of

Best Answer

Related questions

Talk to Our counsellor