Chapter 25, Problem 72SCQ

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

Collision of an electron and a positron results in formation of two γ rays. In the process, their masses are convened completely into energy. (a) Calculate the energy evolved from the annihilation of an electron and a positron, in kilojoules per mole. (b) Using Planck’s equation (Equation 6.2), determine the frequency of the γ rays emitted in this process.

(a)

Interpretation Introduction

Interpretation:

Energy evolved from the annihilation of an electron and a positron should be calculated.

Concept introduction:

Mass defect can be calculated using following formula,

Δm = mR- mp

Here

Δm is mass defect for the process

mR is mass for the reactants

mp is mass of the products

Conversion of amu to kJ

1amu=931.5MeV×(1.602×1010kJ)1MeV

Explanation

Annihilation of an electron and positron gives two gamma rays.

â€‚Â eâˆ’10+Î²+10â†’2Î³

Mass of electron is 0.00054â€‰amu.

Mass of positron is 0.00054â€‰amu .

â€‚Â Î”mÂ =Â mR-Â mp

Here

â€‚Â Î”m is mass defect for the process

â€‚Â mR is mass for the reactants

â€‚Â mp is mass of the products

0.00054â€‰amuâ€‰forâ€‰mR0.00054â€‰amuâ€‰forâ€‰mÎ²0â€‰forâ€‰Î³

Substitute the values in the equation

Î”mÂ =Â (0.0054Â +Â 0

(b)

Interpretation Introduction

Interpretation:

Frequency of the gamma rays should be calculated using Planck’s equation.

Concept introduction:

Planck’s equation is given as:

E=where, ν=frequencyh=Planck’sconstant

The energy increases as the wavelength of the light decreases.

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