Respuesta :
Energy required to break the nucleus= 9.373×10¹³Joules
How much energy must be supplied?
Given:
Mass = 120gm
mass of nucleus of Al-27= 26.9815386
Solution:
Number of Protons in Al-27 = 13
and the no. of Neutrons in Al-27 = (27-13) = 14
Δm = (no. of P × mass of Proton) + (no. of N × mass of neutron)
- [mass of Al-27 nucleus]
Δm = { 13× 1.0072765 amu + 14 × 1.0086649] - -[26.9815386 amu]
Δm = 0.234364 amu.
Energy = binding Energy = mc^2 where C=3x10⁸m/s
E = (0.2343645 amu x 1.661X10⁻²⁷kg) x (3 × 10⁸m/s²)
E = 3.5 X 10⁻¹¹Joules
Number of Nucleus in 120g= (Mass/Molar mass)×6.022×10^23 nucleus/mol
no. of Nucleus = (120g/26.98gm/mol)x 6.022X 10^23 Nucleus/mol
no. of Nucleus = 2.678 X 10²⁴ nucleus
E = No of Nucleus × E per nuclei
E= 2.678 × 10²⁴ Nucleus × 3.5 × 10⁻"J
E =9.373 X 10¹³joules
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Note: The question is not complete.
Complete Question: How Much Energy Must Be Supplied To Break A Single Aluminum-27 Nucleus Into Separated Protons And Neutrons If An Aluminium-27 Atom Has Mass Of 26.9815386?
