Answer:
m(94Be) = 9.012 182 24 amuN(protons) = 4N(neutrons) = 9 – 4 = 5m(protons) = (4 protons) (1.007 276 47 amu/proton) = 4.029 105 88 amum(neutrons) = (5 neutrons) (1.008 664 90 amu/neutron) = 5.043 324 50 amum(protons) + m(neutrons) = 4.029 105 88 amu + 5.043 324 50 amu = 9.072 430 38 amuMass defect m = 9.072 430 38 amu – 9.012 182 24 amu = 0.060 248 14 amum = (0.060 248 14 amu/nucleus) (1.6605 10-27kg/amu) = 1.0004 10-28 kg/nucleusE =mc2= (1.0004 10-28 kg/nucleus) (3.00 108m/s)2= 9.0038 10-12J/nucleusN(nucleons)= N(protons)+N(neutrons) = 9E =(9.0038 10-12J/nucleus) (1 nucleus/9nucleons) =1.0004 10-12J/nucleon2.
The mass defect is 0.059 948 14 amu and the Binding energy/nucleon of the nuclide is 9.66 × 10⁻¹² J/nucleon.
It is expressed as
[tex]\Delta M = (Zm_{P} + Nm_{n}) - M_{A}[/tex]
where
[tex]\Delta M[/tex] = Mass defect
[tex]M_{A}[/tex] = Mass of nucleus
[tex]m_{P}[/tex] = mass of proton
[tex]m_{n}[/tex] = mass of electron
Z = Number of proton
N = Number of neutrons
Now put the values in above formula we get
[tex]\Delta M = (Zm_{P} + Nm_{n}) - M_{A}[/tex]
[tex]\Delta M = 4(m_{p}) + 5 (m_{n}) - M_{A}[/tex]
= 4 (1.007 276 47 amu) + 5 (1.008 664 90 amu) - 9.012 182 24 amu
= 4.029 105 88 amu + 5.043 324 50 amu - 9.012 182 24 amu
= 9.072 130 38 amu - 9.012 182 24 amu
= 0.059 948 14 amu
It is expressed as
ΔE = Δmc²
where
ΔE = Binding energy
ΔM = change in mass
c = speed of light
Now put the values in above formula we get
ΔE = Δmc²
= (0.59 984 14 amu) (1.6605 × 10⁻²⁷ kg/amu) (3 × 10⁸) (m/s)²
Nucleons = 8.96 × 10⁻¹¹
Binding energy/nucleon = 9.66 × 10⁻¹² J/nucleon
Thus from the above conclusion we can say that The mass defect is 0.059 948 14 amu and the Binding energy/nucleon of the nuclide is 9.66 × 10⁻¹² J/nucleon.
Learn more about the Binding energy here: https://brainly.com/question/23020604
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