Answer:
[tex]1.85\cdot 10^{-11}J[/tex]
Explanation:
The binding energy of a nucleus is given by
[tex]\Delta E=c^2 \Delta m[/tex] (1)
where
[tex]c=3.0\cdot 10^8 m/s[/tex] is the speed of light
[tex]\Delta m[/tex] is the mass defect, which is the difference between the sum of the masses of the individual nucleons and the total mass of the nucleus.
Here the mass of the nucleus is
[tex]m(^{15}_7N) = 15.000109 u[/tex]
While the mass of the nucleons is:
[tex]m(p)=1.007825 u[/tex] (mass of the proton)
[tex]m(n)=1.008665u[/tex] (mass of the neutron)
The nucleus [tex]^{15}_7N[/tex] containes 7 protons (atomic number) and 15 nucleons (mass number), which means that the number of neutrons is
[tex]n=15-7=8[/tex]
So the mass defect is:
[tex]\Delta m=(7m(p)+8m(n))-m(^{15}_7N)=\\(7\cdot 1.007825+8\cdot 1.008665)-15.000109)=0.123986u[/tex]
1 atomic mass unit is
[tex]1 u = 1.66054\cdot 10^{-27}kg[/tex]
So the mass defect in kilograms is
[tex]\Delta m=(0.123986)(1.66054\cdot 10^{-27})=2.059\cdot 10^{-28}kg[/tex]
Finally, we can use eq(1) to find the binding energy:
[tex]\Delta E=(3\cdot 10^8)^2 (2.059\cdot 10^{-28})=1.85\cdot 10^{-11}J[/tex]
The binding energy per nucleon for a [tex]^{15}_{7}N[/tex] nucleus is 7.7 MeV. It was calculated with the mass of the neutral atom [tex]^{15}_{7}N[/tex] (15.000109 u), the mass of proton ¹H (1.007825 u), and the mass of neutron (1.008665 u).
We can calculate the binding energy per nucleon with the following equation:
[tex] B = \frac{\Delta mc^{2}}{A} [/tex]
Where:
c: is the speed of light
A: is the number of nucleons in the nucleus of [tex]^{15}_{7}N[/tex] = 15
We know that 1 u = 931.5 MeV/c², so the binding energy is:
[tex] B = (p*m_{p} + n*m_{n} - m_{^{15}_{7}N})\frac{931.5 MeV/u}{A} [/tex] (1)
Where:
p: is the number of protons = 7
n: is the number of neutrons = A - p = 15 - 7 = 8
[tex]m_{p}[/tex]: is the proton's mass = 1.007825 u
[tex]m_{n}[/tex]: is the neutron's mass = 1.008665 u
[tex] m_{^{15}_{7}N}[/tex] = 15.000109 u
By entering the above values into equation (1), we have:
[tex]B = (7*1.007825 u + 8*1.008665 u - 15.000109 u)\frac{931.5 MeV/u}{15} =7.7 \:MeV/nucleon[/tex]
Therefore, the binding energy per nucleon is 7.7 MeV.
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