Problem: You may want to reference (Pages 920 - 922)Section 21.6 while completing this problemHow much energy must be supplied to break a single aluminum-27 nucleus into separated protons and neutrons if an aluminum-27 atom has a mass of 26.9815386 amu? How much energy is required for 100.0 g of aluminum-27? (The mass of an electron is 5.485799 10-4 amu, the mass of a proton is 1.0072765 amu, and the mass of a neutron is 1.0086649 amu.)

FREE Expert Solution

For 27Al:

Atomic number = number or protons = 13

Atomic mass = 27 amu


27 = 13 + # of neutrons

# of neutrons = 27-13

# of neutrons = 14


The nuclear reaction is: 

A1327l 14  n01+13 p11


(1): Calculate the mass defect (Δm).

m=products-reactants

m=[14(mass n01)+13(mass p11)]-mass  1327Alm=[14(1.0086649 amu)+13(1.0072765 amu)]-(26.9815386 amu)

1 amu = 1.6606x10-27 kg

m=0.2343645 amu×1.6606×10-27 kg1 amu  

Δm = 3.892x10-28 kg


(2) Calculate the energy used (E).


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Problem Details

You may want to reference (Pages 920 - 922)Section 21.6 while completing this problem

How much energy must be supplied to break a single aluminum-27 nucleus into separated protons and neutrons if an aluminum-27 atom has a mass of 26.9815386 amu? How much energy is required for 100.0 g of aluminum-27? (The mass of an electron is 5.485799 10-4 amu, the mass of a proton is 1.0072765 amu, and the mass of a neutron is 1.0086649 amu.)

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