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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Chapter 28, Problem 45P
To determine

The ionization energies of L, M and N shell.

Expert Solution & Answer
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Answer to Problem 45P

The ionization energies of L, M and N shell is 11.7keV , 10 keV , 2,3 keV respectively.

Explanation of Solution

Section1:

To determine: The ionization energy of the L-shell.

Answer: The ionization energy of the L-shell is 11.7keV .

Explanation:

Given Info: The ionization energy of K-shell is 69.5keV .

Formula to calculate the ionization energy of the L-shell is,

EL=Ek+hcλ3

  • EL is ionization energy of the L-shell,
  • EK is ionization energy of the K-shell,
  • h is the Planck’s Constant
  • c is the speed of light
  • λ3 is the wavelength

From unit conversion,

1keV=1.6×1016J

Substitute 6.63×1034J-s for h , 3×108m/s for c , 0.0215nm for λ3 , (69.5keV) for EK to find EL ,

EL=(69.5keV)+[(6.63×1034J-s)(3×108m/s)(0.0215nm)(1×109m1nm)(1keV1.6×1016J)]=11.7keV

Thus, ionization energy of the L-shell electron is 11.7keV .

Section 2:

To determine: The ionization energy of the M-shell.

Answer: The ionization energy of the M-shell is 10keV .

Explanation:

Given Info: The ionization energy of K-shell is 69.5keV .

Formula to calculate the ionization energy of the M-shell is,

EM=Ek+hcλ2

  • EM is ionization energy of the M-shell,
  • EK is ionization energy of the K-shell,
  • h is the Planck’s Constant
  • c is the speed of light
  • λ2 is the wavelength

From unit conversion,

1keV=1.6×1016J

Substitute 6.63×1034J-s for h , 3×108m/s for c , 0.0209nm for λ2 , (69.5keV) for EK to find EM ,

EM=(69.5keV)+[(6.63×1034J-s)(3×108m/s)(0.0209nm)(1×109m1nm)(1keV1.6×1016J)]=10keV

Thus, ionization energy of the M-shell electron is 10keV .

Section3:

To determine: The ionization energy of the N-shell.

Answer: The ionization energy of the N-shell is 2.3keV .

Explanation:

Given Info: The ionization energy of K-shell is 69.5keV .

Formula to calculate the ionization energy of the N-shell is,

EN=Ek+hcλ1

  • EN is ionization energy of the L-shell,
  • EK is ionization energy of the K-shell,
  • h is the Planck’s Constant
  • c is the speed of light
  • λ1 is the wavelength

From unit conversion,

1keV=1.6×1016J

Substitute 6.63×1034J-s for h , 3×108m/s for c , 0.0185nm for λ1 , (69.5keV) for EK to find EN ,

EN=(69.5keV)+[(6.63×1034J-s)(3×108m/s)(0.0185nm)(1×109m1nm)(1keV1.6×1016J)]=2.3keV

Thus, ionization energy of the N-shell electron is 2.3keV .

Conclusion:

Therefore, the ionization energies of L, M and N shell is 11.7keV , 10 keV , 2,3 keV respectively.

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