Saved Bookmarks
| 1. |
Calculate the mass defect and the binding energy per nucleon of the ""_(47)^(108)Ag nucleus. [atomic mass of Ag = 107.905949] |
|
Answer» SOLUTION :MASS of proton, `m_(p) = 1.007825 amu` Mass of neutron, `m_(m) = 1.008665 amu` Mass defect, `Delta m = Zm_(p) + Zm_(N) - M_(N)` ` = 47 XX 1.007825 + 61 xx 1.00865 - 107.905949` ` = 108.89634 - 107.905949` `Delta m = 0.990391` u BINDING energy per nucleon of the `""_(47)^(108)"Ag"` NUCLEUS `overline(B.E) = (Delta m x 931)/(A) = (0.990391 xx 931)/(108)` ` = (922.05421)/(108) = 8.537` `overline(B.E) = 8.5 (MeV)/(A)` |
|