Chapter 25, Problem 33PS

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
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

5H11â€‰â€‰+â€‰â€‰5n01â†’B510

Masses of the individual nucleons

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

The mass of the nucleus =10.01294g/mol

The mass defect:

Î”m=M-m

Î”m=10.01294-10.0825Â =Â 0.06956g/molÎ”m=Â 0.06956Ã—10âˆ’3kg/mol

Binding energy:

Eb=Î”mÃ—(c)2=0.06956Ã—10âˆ’3kg/molÃ—(3Ã—108m/s)2=6.260Ã—1012J=6.260Ã—109kJ

The binding energy for boron-10=6.260Ã—108kJ

Binding energy per nucleon =Totalâ€‰â€‰bindingâ€‰â€‰energyâ€‰â€‰ofâ€‰â€‰nucleusNumberâ€‰â€‰ofâ€‰â€‰nucleonsâ€‰â€‰inâ€‰â€‰theâ€‰â€‰nucleus

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

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