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

Question

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Answer 1

Question

Chapter 29, Problem 38P

(a)

To determine

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

(a)

Expert Solution

Answer to Problem 38P

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

Explanation of Solution

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

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card), 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 69.5 keV, so the energy of the K shell is −69.5 keV

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 6.626×10−34 J⋅s for h, 3×108 m/s for c, and 0.0185 nm for λ1 in the equation (I) to find E1.

E1=(6.626×10−34 J⋅s)(3×108 m/s)(0.0185 nm)=19.878×10−26 J⋅m(6.242×1018 eV1 J)(0.0185 nm)=1240 eV⋅nm0.0185 nm=67.03×103 eV

Convert the energy of the photon into kilo electron volt.

E1=67.03×103 eV(10−3 keV1 eV)=67.03 keV

Substitute 6.626×10−34 J⋅s for h, 3×108 m/s for c, and 0.0209 nm for λ2 in the equation (I) to find E2.

E2=(6.626×10−34 J⋅s)(3×108 m/s)(0.0209 nm)=19.878×10−26 J⋅m(6.242×1018 eV1 J)(0.0209 nm)=1240 eV⋅nm0.0209 nm=59.3×103 eV

Convert the energy of the photon into kilo electron volt.

E2=59.3×103 eV(10−3 keV1 eV)=59.3 keV

Substitute 6.626×10−34 J⋅s for h, 3×108 m/s for c, and 0.0215 nm for λ3 in the equation (I) to find E3.

E3=(6.626×10−34 J⋅s)(3×108 m/s)(0.0215 nm)=19.878×10−26 J⋅m(6.242×1018 eV1 J)(0.0215 nm)=1240 eV⋅nm0.0215 nm=57.7×103 eV

Convert the energy of the photon into kilo electron volt.

E3=57.7×103 eV(10−3 keV1 eV)=57.7 keV

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

λ(nm)	Photon energy (keV)	transition	Energy of level	level
λ1=0.0185	67.03	n=4→1	−69.5 keV+67.03 keV=−2.47 keV	N
λ2=0.0209	59.3	n=3→1	−69.5 keV+59.3 keV=−10.2 keV	M
λ3=0.0215	57.7	n=2→1	−69.5 keV+57.7 keV=−11.8 keV	L

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

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

(b)

To determine

Draw the diagram of the transition.

(b)

Expert Solution

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)	transition	Energy of level	level
λ1=0.0185	67.03	n=4→1	−69.5 keV+67.03 keV=−2.47 keV	N
λ2=0.0209	59.3	n=3→1	−69.5 keV+59.3 keV=−10.2 keV	M
λ3=0.0215	57.7	n=2→1	−69.5 keV+57.7 keV=−11.8 keV	L

Conclusion:

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

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Answer 2

Question

Chapter 29, Problem 38P

(a)

To determine

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

(a)

Expert Solution

Answer to Problem 38P

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

Explanation of Solution

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

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card), 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 69.5 keV, so the energy of the K shell is −69.5 keV

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 6.626×10−34 J⋅s for h, 3×108 m/s for c, and 0.0185 nm for λ1 in the equation (I) to find E1.

E1=(6.626×10−34 J⋅s)(3×108 m/s)(0.0185 nm)=19.878×10−26 J⋅m(6.242×1018 eV1 J)(0.0185 nm)=1240 eV⋅nm0.0185 nm=67.03×103 eV

Convert the energy of the photon into kilo electron volt.

E1=67.03×103 eV(10−3 keV1 eV)=67.03 keV

Substitute 6.626×10−34 J⋅s for h, 3×108 m/s for c, and 0.0209 nm for λ2 in the equation (I) to find E2.

E2=(6.626×10−34 J⋅s)(3×108 m/s)(0.0209 nm)=19.878×10−26 J⋅m(6.242×1018 eV1 J)(0.0209 nm)=1240 eV⋅nm0.0209 nm=59.3×103 eV

Convert the energy of the photon into kilo electron volt.

E2=59.3×103 eV(10−3 keV1 eV)=59.3 keV

Substitute 6.626×10−34 J⋅s for h, 3×108 m/s for c, and 0.0215 nm for λ3 in the equation (I) to find E3.

E3=(6.626×10−34 J⋅s)(3×108 m/s)(0.0215 nm)=19.878×10−26 J⋅m(6.242×1018 eV1 J)(0.0215 nm)=1240 eV⋅nm0.0215 nm=57.7×103 eV

Convert the energy of the photon into kilo electron volt.

E3=57.7×103 eV(10−3 keV1 eV)=57.7 keV

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

λ(nm)	Photon energy (keV)	transition	Energy of level	level
λ1=0.0185	67.03	n=4→1	−69.5 keV+67.03 keV=−2.47 keV	N
λ2=0.0209	59.3	n=3→1	−69.5 keV+59.3 keV=−10.2 keV	M
λ3=0.0215	57.7	n=2→1	−69.5 keV+57.7 keV=−11.8 keV	L

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

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

(b)

To determine

Draw the diagram of the transition.

(b)

Expert Solution

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)	transition	Energy of level	level
λ1=0.0185	67.03	n=4→1	−69.5 keV+67.03 keV=−2.47 keV	N
λ2=0.0209	59.3	n=3→1	−69.5 keV+59.3 keV=−10.2 keV	M
λ3=0.0215	57.7	n=2→1	−69.5 keV+57.7 keV=−11.8 keV	L

Conclusion:

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

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Answer 3

Question

Chapter 29, Problem 38P

(a)

To determine

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

(a)

Expert Solution

Answer to Problem 38P

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

Explanation of Solution

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

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card), 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 69.5 keV, so the energy of the K shell is −69.5 keV

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 6.626×10−34 J⋅s for h, 3×108 m/s for c, and 0.0185 nm for λ1 in the equation (I) to find E1.

E1=(6.626×10−34 J⋅s)(3×108 m/s)(0.0185 nm)=19.878×10−26 J⋅m(6.242×1018 eV1 J)(0.0185 nm)=1240 eV⋅nm0.0185 nm=67.03×103 eV

Convert the energy of the photon into kilo electron volt.

E1=67.03×103 eV(10−3 keV1 eV)=67.03 keV

Substitute 6.626×10−34 J⋅s for h, 3×108 m/s for c, and 0.0209 nm for λ2 in the equation (I) to find E2.

E2=(6.626×10−34 J⋅s)(3×108 m/s)(0.0209 nm)=19.878×10−26 J⋅m(6.242×1018 eV1 J)(0.0209 nm)=1240 eV⋅nm0.0209 nm=59.3×103 eV

Convert the energy of the photon into kilo electron volt.

E2=59.3×103 eV(10−3 keV1 eV)=59.3 keV

Substitute 6.626×10−34 J⋅s for h, 3×108 m/s for c, and 0.0215 nm for λ3 in the equation (I) to find E3.

E3=(6.626×10−34 J⋅s)(3×108 m/s)(0.0215 nm)=19.878×10−26 J⋅m(6.242×1018 eV1 J)(0.0215 nm)=1240 eV⋅nm0.0215 nm=57.7×103 eV

Convert the energy of the photon into kilo electron volt.

E3=57.7×103 eV(10−3 keV1 eV)=57.7 keV

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

λ(nm)	Photon energy (keV)	transition	Energy of level	level
λ1=0.0185	67.03	n=4→1	−69.5 keV+67.03 keV=−2.47 keV	N
λ2=0.0209	59.3	n=3→1	−69.5 keV+59.3 keV=−10.2 keV	M
λ3=0.0215	57.7	n=2→1	−69.5 keV+57.7 keV=−11.8 keV	L

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

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

(b)

To determine

Draw the diagram of the transition.

(b)

Expert Solution

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)	transition	Energy of level	level
λ1=0.0185	67.03	n=4→1	−69.5 keV+67.03 keV=−2.47 keV	N
λ2=0.0209	59.3	n=3→1	−69.5 keV+59.3 keV=−10.2 keV	M
λ3=0.0215	57.7	n=2→1	−69.5 keV+57.7 keV=−11.8 keV	L

Conclusion:

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

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Answer 4

Question

Chapter 29, Problem 38P

(a)

To determine

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

(a)

Expert Solution

Answer to Problem 38P

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

Explanation of Solution

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

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card), 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 69.5 keV, so the energy of the K shell is −69.5 keV

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 6.626×10−34 J⋅s for h, 3×108 m/s for c, and 0.0185 nm for λ1 in the equation (I) to find E1.

E1=(6.626×10−34 J⋅s)(3×108 m/s)(0.0185 nm)=19.878×10−26 J⋅m(6.242×1018 eV1 J)(0.0185 nm)=1240 eV⋅nm0.0185 nm=67.03×103 eV

Convert the energy of the photon into kilo electron volt.

E1=67.03×103 eV(10−3 keV1 eV)=67.03 keV

Substitute 6.626×10−34 J⋅s for h, 3×108 m/s for c, and 0.0209 nm for λ2 in the equation (I) to find E2.

E2=(6.626×10−34 J⋅s)(3×108 m/s)(0.0209 nm)=19.878×10−26 J⋅m(6.242×1018 eV1 J)(0.0209 nm)=1240 eV⋅nm0.0209 nm=59.3×103 eV

Convert the energy of the photon into kilo electron volt.

E2=59.3×103 eV(10−3 keV1 eV)=59.3 keV

Substitute 6.626×10−34 J⋅s for h, 3×108 m/s for c, and 0.0215 nm for λ3 in the equation (I) to find E3.

E3=(6.626×10−34 J⋅s)(3×108 m/s)(0.0215 nm)=19.878×10−26 J⋅m(6.242×1018 eV1 J)(0.0215 nm)=1240 eV⋅nm0.0215 nm=57.7×103 eV

Convert the energy of the photon into kilo electron volt.

E3=57.7×103 eV(10−3 keV1 eV)=57.7 keV

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

λ(nm)	Photon energy (keV)	transition	Energy of level	level
λ1=0.0185	67.03	n=4→1	−69.5 keV+67.03 keV=−2.47 keV	N
λ2=0.0209	59.3	n=3→1	−69.5 keV+59.3 keV=−10.2 keV	M
λ3=0.0215	57.7	n=2→1	−69.5 keV+57.7 keV=−11.8 keV	L

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

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

(b)

To determine

Draw the diagram of the transition.

(b)

Expert Solution

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)	transition	Energy of level	level
λ1=0.0185	67.03	n=4→1	−69.5 keV+67.03 keV=−2.47 keV	N
λ2=0.0209	59.3	n=3→1	−69.5 keV+59.3 keV=−10.2 keV	M
λ3=0.0215	57.7	n=2→1	−69.5 keV+57.7 keV=−11.8 keV	L

Conclusion:

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

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Answer 5

Chapter 29, Problem 38P

(a)

To determine

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

(a)

Expert Solution

Answer to Problem 38P

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

Explanation of Solution

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

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card), 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 69.5 keV, so the energy of the K shell is −69.5 keV

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 6.626×10−34 J⋅s for h, 3×108 m/s for c, and 0.0185 nm for λ1 in the equation (I) to find E1.

E1=(6.626×10−34 J⋅s)(3×108 m/s)(0.0185 nm)=19.878×10−26 J⋅m(6.242×1018 eV1 J)(0.0185 nm)=1240 eV⋅nm0.0185 nm=67.03×103 eV

Convert the energy of the photon into kilo electron volt.

E1=67.03×103 eV(10−3 keV1 eV)=67.03 keV

Substitute 6.626×10−34 J⋅s for h, 3×108 m/s for c, and 0.0209 nm for λ2 in the equation (I) to find E2.

E2=(6.626×10−34 J⋅s)(3×108 m/s)(0.0209 nm)=19.878×10−26 J⋅m(6.242×1018 eV1 J)(0.0209 nm)=1240 eV⋅nm0.0209 nm=59.3×103 eV

Convert the energy of the photon into kilo electron volt.

E2=59.3×103 eV(10−3 keV1 eV)=59.3 keV

Substitute 6.626×10−34 J⋅s for h, 3×108 m/s for c, and 0.0215 nm for λ3 in the equation (I) to find E3.

E3=(6.626×10−34 J⋅s)(3×108 m/s)(0.0215 nm)=19.878×10−26 J⋅m(6.242×1018 eV1 J)(0.0215 nm)=1240 eV⋅nm0.0215 nm=57.7×103 eV

Convert the energy of the photon into kilo electron volt.

E3=57.7×103 eV(10−3 keV1 eV)=57.7 keV

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

λ(nm)	Photon energy (keV)	transition	Energy of level	level
λ1=0.0185	67.03	n=4→1	−69.5 keV+67.03 keV=−2.47 keV	N
λ2=0.0209	59.3	n=3→1	−69.5 keV+59.3 keV=−10.2 keV	M
λ3=0.0215	57.7	n=2→1	−69.5 keV+57.7 keV=−11.8 keV	L

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

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

(b)

To determine

Draw the diagram of the transition.

(b)

Expert Solution

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)	transition	Energy of level	level
λ1=0.0185	67.03	n=4→1	−69.5 keV+67.03 keV=−2.47 keV	N
λ2=0.0209	59.3	n=3→1	−69.5 keV+59.3 keV=−10.2 keV	M
λ3=0.0215	57.7	n=2→1	−69.5 keV+57.7 keV=−11.8 keV	L

Conclusion:

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

Answer 6

Chapter 29, Problem 38P

(a)

To determine

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

(a)

Expert Solution

Answer to Problem 38P

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

Explanation of Solution

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

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card), 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 69.5 keV, so the energy of the K shell is −69.5 keV

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 6.626×10−34 J⋅s for h, 3×108 m/s for c, and 0.0185 nm for λ1 in the equation (I) to find E1.

E1=(6.626×10−34 J⋅s)(3×108 m/s)(0.0185 nm)=19.878×10−26 J⋅m(6.242×1018 eV1 J)(0.0185 nm)=1240 eV⋅nm0.0185 nm=67.03×103 eV

Convert the energy of the photon into kilo electron volt.

E1=67.03×103 eV(10−3 keV1 eV)=67.03 keV

Substitute 6.626×10−34 J⋅s for h, 3×108 m/s for c, and 0.0209 nm for λ2 in the equation (I) to find E2.

E2=(6.626×10−34 J⋅s)(3×108 m/s)(0.0209 nm)=19.878×10−26 J⋅m(6.242×1018 eV1 J)(0.0209 nm)=1240 eV⋅nm0.0209 nm=59.3×103 eV

Convert the energy of the photon into kilo electron volt.

E2=59.3×103 eV(10−3 keV1 eV)=59.3 keV

Substitute 6.626×10−34 J⋅s for h, 3×108 m/s for c, and 0.0215 nm for λ3 in the equation (I) to find E3.

E3=(6.626×10−34 J⋅s)(3×108 m/s)(0.0215 nm)=19.878×10−26 J⋅m(6.242×1018 eV1 J)(0.0215 nm)=1240 eV⋅nm0.0215 nm=57.7×103 eV

Convert the energy of the photon into kilo electron volt.

E3=57.7×103 eV(10−3 keV1 eV)=57.7 keV

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

λ(nm)	Photon energy (keV)	transition	Energy of level	level
λ1=0.0185	67.03	n=4→1	−69.5 keV+67.03 keV=−2.47 keV	N
λ2=0.0209	59.3	n=3→1	−69.5 keV+59.3 keV=−10.2 keV	M
λ3=0.0215	57.7	n=2→1	−69.5 keV+57.7 keV=−11.8 keV	L

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

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

Answer 7

Chapter 29, Problem 38P

(a)

To determine

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

Answer 8

(a)

Expert Solution

Answer to Problem 38P

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

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

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

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card), 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 69.5 keV, so the energy of the K shell is −69.5 keV

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 6.626×10−34 J⋅s for h, 3×108 m/s for c, and 0.0185 nm for λ1 in the equation (I) to find E1.

E1=(6.626×10−34 J⋅s)(3×108 m/s)(0.0185 nm)=19.878×10−26 J⋅m(6.242×1018 eV1 J)(0.0185 nm)=1240 eV⋅nm0.0185 nm=67.03×103 eV

Convert the energy of the photon into kilo electron volt.

E1=67.03×103 eV(10−3 keV1 eV)=67.03 keV

Substitute 6.626×10−34 J⋅s for h, 3×108 m/s for c, and 0.0209 nm for λ2 in the equation (I) to find E2.

E2=(6.626×10−34 J⋅s)(3×108 m/s)(0.0209 nm)=19.878×10−26 J⋅m(6.242×1018 eV1 J)(0.0209 nm)=1240 eV⋅nm0.0209 nm=59.3×103 eV

Convert the energy of the photon into kilo electron volt.

E2=59.3×103 eV(10−3 keV1 eV)=59.3 keV

Substitute 6.626×10−34 J⋅s for h, 3×108 m/s for c, and 0.0215 nm for λ3 in the equation (I) to find E3.

E3=(6.626×10−34 J⋅s)(3×108 m/s)(0.0215 nm)=19.878×10−26 J⋅m(6.242×1018 eV1 J)(0.0215 nm)=1240 eV⋅nm0.0215 nm=57.7×103 eV

Convert the energy of the photon into kilo electron volt.

E3=57.7×103 eV(10−3 keV1 eV)=57.7 keV

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

λ(nm)	Photon energy (keV)	transition	Energy of level	level
λ1=0.0185	67.03	n=4→1	−69.5 keV+67.03 keV=−2.47 keV	N
λ2=0.0209	59.3	n=3→1	−69.5 keV+59.3 keV=−10.2 keV	M
λ3=0.0215	57.7	n=2→1	−69.5 keV+57.7 keV=−11.8 keV	L

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

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

Answer 13

(b)

To determine

Draw the diagram of the transition.

(b)

Expert Solution

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)	transition	Energy of level	level
λ1=0.0185	67.03	n=4→1	−69.5 keV+67.03 keV=−2.47 keV	N
λ2=0.0209	59.3	n=3→1	−69.5 keV+59.3 keV=−10.2 keV	M
λ3=0.0215	57.7	n=2→1	−69.5 keV+57.7 keV=−11.8 keV	L

Conclusion:

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

Answer 14

(b)

To determine

Draw the diagram of the transition.

Answer 15

(b)

Expert Solution

Answer to Problem 38P

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

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

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

λ(nm)	Photon energy (keV)	transition	Energy of level	level
λ1=0.0185	67.03	n=4→1	−69.5 keV+67.03 keV=−2.47 keV	N
λ2=0.0209	59.3	n=3→1	−69.5 keV+59.3 keV=−10.2 keV	M
λ3=0.0215	57.7	n=2→1	−69.5 keV+57.7 keV=−11.8 keV	L