Explanation The atomic mass of 1 1 H is 1 .0078 amu . The atomic mass of 1 2 H is 2 .01410 amu . The mass of neutron is 1.0087 amu . Number of protons in 1 2 H = 1 Number of neutron in 1 2 H = 1 The stated reaction is, 1 1 H + 1 1 H → 1 2 H+ 1 0 e The mass defect is calculated by the formula, Δ m = [ Mass of 1 2 H nucleus +Mass of positron ] − [ 2×Mass of 1 2 H nucleus ] Δ m = ( [ ( Atomic mass of 1 2 H nucleus − Mass of electron ) + Mass of positron ] − [ 2(Atomic mass of 1 1 H-mass of electron) ] ) Substitute the value of the atomic mass of 1 2 H , the number of protons; the mass of the 1 1 H proton and that of the neutron in the above equation. Δ m = 2 .01410 − [ 2 ( 1.00782 − 5.4858 × 10 − 4 ) ] Δ m = − 0.0044 amu / nucleus The conversion of amu / nucleus to Kg/nucleus is done as, 1 amu = 1.66 × 10 − 27 Kg Therefore, the conversion of − 0.0044 amu / nucleus into kg/nucleus is, − 0.0044 amu / nucleus = ( − 0.0044 × 1.66 × 10 − 27 ) kg/nucleus = - 7 . 3 0 × 1 0 - 3 1 K g / n u c l e u s _ Therefore, the mass defect ( Δ m ) of 1 2 H is - 7 . 3 0 × 1 0 - 3 1 K g / n u c l e u s _

Chemistry with Access Code, Hybrid Edition

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Publisher: CENGAGE L

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Question

Chapter 19, Problem 49E

Interpretation Introduction

Interpretation:

The amount of energy released per gram of hydrogen reacted for the given reaction is to be calculated.

Concept introduction:

The sum of masses of the component nucleons and the actual mass of a nucleus is known as the mass defect and it can be used to calculate the nuclear binding energy.

Expert Solution & Answer

Chapter 19 Solutions

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Francium was discovered as a minor decay product of actinium-227. Write the nuclear equation for the decay of actinium-227 by alpha emission.

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