Question

Calculate the binding energy per nucleon of the helium nucleus 52he. express your answer in millions of electron volts to four significant figures.

Answer 1

The binding energy per nucleon of the helium nucleus (^4He) is calculated by dividing the total binding energy by the number of nucleons. For helium, this value is approximately 0.007594 MeV.

Step-by-step explanation:

The binding energy per nucleon of the helium nucleus (^4He) can be calculated using the equation BE/A, where BE is the total binding energy and A is the number of nucleons. The total binding energy for helium is given by BE = {[2mp +2mn] (4He)} c², where mp is the mass of a proton and mn is the mass of a neutron.

Using the given atomic masses, mp = 1.007825 u and mn = 1.008665 u, the total binding energy is BE = (0.030378 u) c².

To express the binding energy per nucleon in millions of electron volts (MeV), we divide the total binding energy by the number of nucleons A. For helium, A = 4, so the binding energy per nucleon is (0.030378 u) c² / 4 = 0.007594 MeV.

Answer 2

The main formula is given by Eb/nucleon = Eb/ mass of nucleid
as for 52He, the mass is 5
so by applying Einstein's formula Eb=DmC², Eb=binding energy
52He-----------> 2 x 11p + 3 x10n is the equation bilan
so Dm=2 mp + (5-2)mn-mnucleus, mp=mass of proton=1.67 10^-27 kg
mn=mass of neutron=1.67 10^-27 kg
m nucleus= 5
Dm= 2x1.67 10^-27 kg+ 3x1.67 10^-27 kg-5= - 4.9 J
Eb= - 4.9 J x c²= -4.9 x 9 .10^16= - 45 10^16 J
so the answer is Eb /nucleon = Eb/5= -9.10^16 J, but 1eV=1.6 . 10^-19 J
so -9.10^16 J/ 1.6 10^-19= -5.625 10^35 eV

the final answer is Eb /nucleon = -5.625 x10^35 eV