   Chapter 25, Problem 33PS

Chapter
Section
Textbook Problem

Boron has two stable isotopes, 10B and 11B. Calculate the binding energies per mole of nucleons of these two nuclei. The required masses (in g/mol) are H 1 1 = 1.00783, n 0 1 = 1.00867, B 5 10 = 10.01294, and B 5 11 = 11.00931.

Interpretation Introduction

Interpretation:

Binding energy per nucleon of B10 and B11 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 boron-11=11.00931g/mol

Mass of boron-10=10.01294g/mol

Boron-10 has 5 protons and 5 neutrons, and then boron-11 has 5 protons and 6 neutrons.

Binding energy for boron-10:

Equation for boron-10

5H11+5n01B510

Masses of the individual nucleons

5×1.00783=5.039155×1.00867=5.04335=5.03915+5.04335=10.0825g/mol

The mass of the nucleus =10.01294g/mol

The mass defect:

Δm=M-m

Δm=10.01294-10.0825 = 0.06956g/molΔm= 0.06956×103kg/mol

Binding energy:

Eb=Δm×(c)2=0.06956×103kg/mol×(3×108m/s)2=6.260×1012J=6.260×109kJ

The binding energy for boron-10=6.260×108kJ

Binding energy per nucleon =TotalbindingenergyofnucleusNumberofnucleonsinthenucleus

The binding energy per mole of nucleon of boron-10=6

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