Answered: Consider positronium, an atomic-like…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

Consider positronium, an atomic-like system made of an electron and a positron. Find the wavelengths for the transitions n=2 --> 1 and n=3 --> 1. Compare each to the similar transitions for hydrogen.

Expert Solution

Step 1

Reduced mass of positronium, u = mm/m+m = m²/2m = m/2

where m is mass of electron or positron

wavelengths for the transitions n=2 --> 1,

1/ $λ$ = R_H (1/n²_i - 1/n²_f)

n_i = 2 and n_f = 1

1/ $λ$ = R_H (1/2² - 1/1²) = R_H (1/4 - 1/1) = R_H ((4-1)/4) = R_H (3/4)

$λ$ = 4/3R_H

R_H = (R/m_e) x u

where R is Rydberg constant for Hydrogen = 1.097 x 10⁷ m

R_H = R/m_e x m/2 = R/2 = 1.097 x 10⁷ m/2 = 0.5485 x 10⁷ m

$λ$ = 4/3R_H = 4 / 0.5485 x 10⁷ mx3 = 4/1.6455 x 10⁷ = 2.431 x 10^-⁷ m = 243.1 x 10^-9 m = 243.1 nm

wavelengths for the transitions n=3 --> 1,

1/ $λ$ = R_H (1/n²_i - 1/n²_f)

n_i = 3 and n_f = 1

1/ $λ$ = R_H (1/3² - 1/1²) = R_H (1/9 - 1/1) = R_H ((9-1)/9) = R_H (8/9)

$λ$ = 9/8R_H

R_H = (R/m_e) x u

where R is Rydberg constant for Hydrogen = 1.097 x 10⁷ m

R_H = R/m_e x m/2 = R/2 = 1.097 x 10⁷ m/2 = 0.5485 x 10⁷ m

$λ$ = 9/8R_H = 9 / 0.5485 x 10⁷ mx8 = 9/4.388 x 10⁷ = 2.051 x 10^-⁷ m = 205.1 x 10^-9 m = 205.1 nm

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