ACP COLLEGE PHYS 1101/1102 BUNDLE
ACP COLLEGE PHYS 1101/1102 BUNDLE
11th Edition
ISBN: 9781337685467
Author: SERWAY
Publisher: CENGAGE L
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Question
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Chapter 28, Problem 29P

(a)

To determine

The wavelength of the positronium for the transition n=3 to n=2 .

(a)

Expert Solution
Check Mark

Answer to Problem 29P

The wavelength of the positronium for the transition n=3 to n=2 is 1313nm .

Explanation of Solution

Formula to calculate the wavelength is,

λ=1μZ2(36hc5k)

  • μ is the reduced mass,
  • Z is atomic number,
  • h is the Planck’s constant
  • c is the speed of light
  • k is the Coulomb’s constant

Expression the wavelength for the positronium is,

λp=1μpZp2(36hc5k)

  • μp is the reduced mass for positronium,
  • Zp is atomic number of positronium,

Expression the wavelength for the hydrogen is,

λH=1μHZH2(36hc5k)

  • μH is the reduced mass for hydrogen,
  • ZH is atomic number of hydrogen

Taking the ratio of the wavelength of the positronium to hydrogen,

λpλH=(1μpZp2)(36hc5k)(1μHZH2)(36hc5k)=μHZH2μpZp2

Substituting, me for μH , (me2) for μp , 1 for ZH , 1 for Zp , 656.3nm for λH to find λp ,

λp(656.3nm)=me(1)2(me2)(1)2λp=1313nm

Thus, the wavelength of the positronium is 1313nm .

Conclusion:

Therefore, the wavelength of the positronium is 1313nm .

(b)

To determine

The wavelength of the singly ionized helium for the transition n=3 to n=2 .

(b)

Expert Solution
Check Mark

Answer to Problem 29P

The wavelength of the singly ionized helium for the transition n=3 to n=2 is 164.1nm .

Explanation of Solution

Formula to calculate the wavelength is,

λ=1μZ2(36hc5k)

  • μ is the reduced mass,
  • Z is atomic number,
  • h is the Planck’s constant
  • c is the speed of light
  • k is the Coulomb’s constant

Expression the wavelength for the helium is,

λHe=1μHeZHe2(36hc5k)

  • μHe is the reduced mass for Helium
  • ZHe is atomic number of Helium

Expression the wavelength for the hydrogen is,

λH=1μHZH2(36hc5k)

  • μH is the reduced mass for hydrogen,
  • ZH is atomic number of hydrogen

Taking the ratio of the wavelength of the Helium to hydrogen,

λHeλH=(1μHeZHe2)(36hc5k)(1μHZH2)(36hc5k)=μHZH2μHeZHe2

Substituting, me for μH , me for μHe , 1 for ZH , 2 for Zp , 656.3nm for λH to find λHe ,

λHe(656.3nm)=me(1)2(me)(2)2λHe=164.1nm

Thus, the wavelength of the Helium is 164.1nm .

Conclusion:

Therefore, the wavelength of the Helium is 164.1nm

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