Calculate the nuclear binding energy of one Potassium-35 atom if the measured atomic mass of is 34.988011 amu?

1 Answer
Truong-Son N.
May 15, 2018

I get about #"7.9653 MeV/nucleon"#.

Well, I guess I'll have to provide you the atomic mass of #""_(1)^(1)"H"# of #"1.007825 amu"# and neutron mass #"1.008664 amu"#...

From this, the theoretical atomic mass, if there was zero binding energy, would have been:

#""_(19)^(35) "K"#:

#M_r = 19 cancel("protons + electrons") xx "1.007825 amu"/(cancel(""_(1)^(1) "H atom")) + (35 - 19 cancel"neutrons") xx "1.008664 amu/"cancel"neutron"#

#=# #"35.287299 amu"#

So, the mass defect is:

#"35.287299 amu" - "34.988011 amu" = "0.299288 amu"#

As a result, knowing that #"1 amu"# of mass defect involves about #"931.5 MeV"# of binding energy,

#E_"binding" = "931.5 MeV"/cancel"amu" xx 0.299288 cancel"amu" = "278.79 MeV"#

But I assume you want it in terms of #"MeV/nucleon"#, so...

#color(blue)(E_"binding") = "278.79 MeV"/("35 nucleons") ~~ color(blue)("7.9653 MeV/nucleon")#

(The actual value is about #"7.965676 MeV/nucleon"#.)

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