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Chapter 29, Problem 38P

(a)

To determine

The ionization energies of the L, M and N shells.

(a)

Expert Solution
Check Mark

Answer to Problem 38P

The ionization energies for the L shell is 11.8keV, M shell is 10.2keV and N shells is 2.47keV

Explanation of Solution

The figure 1 shows the transitions from higher levels N shell down to K shell.

Bundle: Principles of Physics: A Calculus-Based Text, 5th + WebAssign Printed Access Card for Serway/Jewett's Principles of Physics: A Calculus-Based Text, 5th Edition, Multi-Term, Chapter 29, Problem 38P

The K series includes the transition from higher levels down to the K shell (n=1). Transition from higher state produces photons of higher energy.

The ionization energy for the K shell is 69.5keV, so the energy of the K shell is 69.5keV

Write the expression for energy of the photon.

  E=hcλ        (I)

Here, E is the energy of the photon, h is the plank’s constant, c is the velocity of light, and λ is the wavelength.

Conclusion:

Substitute 6.626×1034Js for h, 3×108m/s for c, and 0.0185nm for λ1 in the equation (I) to find E1.

  E1=(6.626×1034Js)(3×108m/s)(0.0185nm)=19.878×1026Jm(6.242×1018eV1J)(0.0185nm)=1240eVnm0.0185nm=67.03×103eV

Convert the energy of the photon into kilo electron volt.

  E1=67.03×103eV(103keV1eV)=67.03keV

Substitute 6.626×1034Js for h, 3×108m/s for c, and 0.0209nm for λ2 in the equation (I) to find E2.

  E2=(6.626×1034Js)(3×108m/s)(0.0209nm)=19.878×1026Jm(6.242×1018eV1J)(0.0209nm)=1240eVnm0.0209nm=59.3×103eV

Convert the energy of the photon into kilo electron volt.

  E2=59.3×103eV(103keV1eV)=59.3keV

Substitute 6.626×1034Js for h, 3×108m/s for c, and 0.0215nm for λ3 in the equation (I) to find E3.

  E3=(6.626×1034Js)(3×108m/s)(0.0215nm)=19.878×1026Jm(6.242×1018eV1J)(0.0215nm)=1240eVnm0.0215nm=57.7×103eV

Convert the energy of the photon into kilo electron volt.

  E3=57.7×103eV(103keV1eV)=57.7keV

Similarly the table 1 shows the ionization energy for the shells.

λ(nm)Photon energy (keV)transitionEnergy of levellevel
λ1=0.018567.03n=4169.5keV+67.03keV=2.47keVN
λ2=0.020959.3n=3169.5keV+59.3keV=10.2keVM
λ3=0.021557.7n=2169.5keV+57.7keV=11.8keVL

The ionization energy for the K shell is 69.5keV, so the ionization energies for other shell.

Therefore, the ionization energies for the L shell is 11.8keV, M shell is 10.2keV and N shells is 2.47keV

(b)

To determine

Draw the diagram of the transition.

(b)

Expert Solution
Check Mark

Answer to Problem 38P

The table 1 shows the transition of the X-ray spectrum of tungsten element.

Explanation of Solution

From (a) The table shows the transition of the X ray spectrum..

λ(nm)Photon energy (keV)transitionEnergy of levellevel
λ1=0.018567.03n=4169.5keV+67.03keV=2.47keVN
λ2=0.020959.3n=3169.5keV+59.3keV=10.2keVM
λ3=0.021557.7n=2169.5keV+57.7keV=11.8keVL

Conclusion:

Therefore, the table 1 shows the transition of the X-ray spectrum of tungsten element..

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Students have asked these similar questions
The K series of the discrete x-ray spectrum of tungsten contains wavelengths of 0.018 5 nm, 0.020 9 nm, and 0.021 5 nm. The K-shell ionization energy is 69.5 keV. (a) Determine the ionization energies of the L, M, and N shells. L shell _keV M shell keV N shell _keV
The K series of the discrete spectrum of tungsten contains wavelengths of 0.018 5 nm, 0.020 9 nm, and 0.021 5 nm. The K-shell ionization energy is 69.5 keV. Determine the ionization energies of the L, M, and N shells.
The K series of the discrete spectrum of tungsten contains wavelengths of 0.0185 nm, 0.0209 nm, and 0.0215 nm. The K-shell ionization energy is 69.5 keV. Determine the ionization energies of the L, M, and N shells. kev EL = 11.83 kev 2.4 Your incorrect answer may have resulted from roundoff error. Make sure you keep extra significant figures in intermediate steps of your calculation. ke EN EM = 10.17 %D

Chapter 29 Solutions

Bundle: Principles of Physics: A Calculus-Based Text, 5th + WebAssign Printed Access Card for Serway/Jewett's Principles of Physics: A Calculus-Based Text, 5th Edition, Multi-Term

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