   Chapter 25, Problem 34PS

Chapter
Section
Textbook Problem

Calculate the binding energy in kilojoules per mole of nucleons of P for the formation of 30P and 31P. The required masses (in g/mol) are H 1 1 = 1.00783, n 0 1 = 1.00867, P 15 30 = 29.97832, and P 15 31 = 30.97376.

Interpretation Introduction

Interpretation:

Binding energy per nucleon of P for the formation of 30P and 31P has to be calculated.

Concept introduction:

Binding energy is the energy required to separate the nucleus of an atom into proton and neutron. Binding energy is given by Eb=Δm×(c)2, where Δm is the mass defect and c is the velocity of light =3×108m/sec-1

Mass defect is the sum of the masses of the individual nucleons that form an atomic nucleus and the mass of the nucleus.

Binding energy per nucleon =TotalbindingenergyofnucleusNumberofnucleonsinthenucleus

Explanation

Given,

Mass of protons =1.00783g/mol

Mass of neutrons =1.00867g/mol

Mass of 30P=29.97832g/mol

Mass of 31P=30.97376g/mol

30P has 15 protons and 15 neutrons, and then 31P has 15 protons and 16 neutrons.

Binding energy for 30P:

Equation for 30P

15H11+15n0130P

The mass of the individual nucleons

15×1.00783=15.1174515×1.00867=15.13005=15.11745+15.13005=30.2475g/mol

The mass of nucleus =29.97832g/mol

The mass defect:

Δm=M-m

Δm=30.247529.97832 = 0.26918g/molΔm=  0.26918×103kg/mol

Binding energy:

Eb=Δm×(c)2=  0.26918×103kg/mol×(3×108m/s)2=2.422×1016J=2.422×1013kJ

The binding energy for 30P=2.422×1013kJ

Binding energy per nucleon =TotalbindingenergyofnucleusNumberofnucleonsinthenucleus

The binding energy per mole of nucleon of 30P=2

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