What is the nuclear binding energy per nucleon of a _21^51 Ni atom? (c = 3.00 ti

Question

What is the nuclear binding energy per nucleon of a _21^51 Ni atom? (c = 3.00 times 10^8 m/s, 1 amu = 1.66054 times 10^-27 kg) A) 1.51 times 10^-10 J/nucleon B) 1.40 times 10^-12 J/nucleon C) 7.42 times 10^-11 J/nucleon D) 1.49 times 10^-10 J/nucleon E) 7.93 times 10^-11 J/nucleon What quantity of energy is released per gram of U-235 based on the following neutron induced fission of U-235? (c = 3.00 times 10^8 m/s) _92^235U + _0^1n rightarrow 2_0^1n + _52^137Te + _40^97Zr A) 7.62 times 10^13 J/g U-235 B) 1.79 times 10^10 J/g U-235 C) 1.79 times 10^13 J/g U-235 D) 7.62 times 10^10 J/g U-235 E) 3.10 times 10^11 J/g U-235 When^235U collides with one neutron, fission occurs. What is one possible set of products? A) four neutrons, ^90Sr, and^139Ce B) four neutrons, ^90Sr, and^142Xe C) four neutrons, ^90Sr, and^139Xe D) four neutrons, ^90Sr, and^141Xe E) four neutrons, ^90Sr, and^140Xe

Explanation / Answer

Q13.

The atomic mass of 58 28 Ni I is 57.935347 amu. Calculate the nuclear binding energy of this

nucleus and the corresponding nuclear binding energy per nucleon

Mass of proton = 1.007825

Mass of neutrón = 1.008665

Total mass = 28*1.007825 + 30*1.008665 = 58.47905 amu

The mass defect:

57.935347-58.47905= -0.543703 amu

Mass = 0.543703/(6.022*10^23) = 9.028*10^-25 g or 9.028*10^-28 kg

Energy released (Einstein E = mC^2)

E = m*C^2 = 9.028*10^-28 * (3*10^8)^2 = 8.125*10^-11 J

Nuclear binding energy = E = 8.125*10^-11 J

Nuclear binding energy per nucleon

Nucleon = neutron + protons = 58

E per nucleon = E / nucleon =( 8.125*10^-11)/( 58) = 1.4008*10^-12 J/nucleon

best answer is B

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