The electronic transition corresponds to line B and C has to be identified.

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Answer 1

Question

Chapter 7, Problem 7.119SP

(a)

Interpretation Introduction

Interpretation:

The electronic transition corresponds to line B and C has to be identified.

(a)

Expert Solution

Explanation of Solution

The given lines are corresponding to $n = 2$ . Line A has longest wavelength or lowest energy transition. This indicates the transition is $3 → 2$ transition. Line B has greater wavelength and lower energy than line C. Therefore, line B corresponds to $4 → 2$ and line C corresponds to $5 → 2$ transition.

(b)

Interpretation Introduction

Interpretation:

From the wavelength of line c, the wavelength of line A and B has to be calculated.

Concept Introduction:

Bohr developed a rule for quantization of energy that could be applicable to the electron of an atom in motion. By using this, he derived a formula for energy levels of electron in H-atom.

$E = -RHn2 n = 1,2,3,......(For hydrogen atom)RH is Rydberg constant (2.179 × 10-18 J).Eis energy level.n is principal quantum number.$

(b)

Expert Solution

Answer to Problem 7.119SP

The wavelength of line A is $41.1nm$

The wavelength of line B is $30.4nm$

Explanation of Solution

The energy of line C is calculated using its wavelength as follows,

$E = hcλ = (6.63 × 10-34J.s)(3.00 × 108m/s)(27.1 × 10-9m) = 7.34 × 10-18J$

The atom in which electronic transition of $5 → 2$ occurs is identified as

$ΔE = hν = Ef - EiΔE = RH Z2 (1ni2 - 1nf2)-7.34 × 10-18J = RH Z2 (1ni2 - 1nf2)-7.34 × 10-18J = (2.18 × 10-18J)Z2(152 - 122)-7.34 × 10-18J = Z2(2.18 × 10-18J52 - 2.18 × 10-18J22)-7.34 × 10-18J = Z2(8.72 × 10-18J - 5.45× 10-17J100)-7.34 × 10-18J = (-4.58 × 10-19)Z2Z2 = 16.0Z = 4$

The energy change and wavelength for transitions of $3 → 2$ and $4 → 2$ are calculated as

$ΔE = RHZ2(1ni2 - 1nf2) = (2.18 × 10-18J) (4)2(132 - 122) = (3.488 × 10-17J) (132 - 122) = (3.488 × 10-17J32 - 3.488 × 10-17J22) = (1.395 × 10-16J- 3.139 × 10-16J 36 ) = -4.84 × 10-18J$

Negative sign is neglected while calculating lambda

$λ = hcΔE = (6.63 × 10-34 J.s)(3.00 × 108m/s)4.84 × 10-18Jλ = 4.11 × 10-8m = 41.1nm 1m = 109nm$

$ΔE = RHZ2(1ni2 - 1nf2) = (2.18 × 10-18J) (4)2(142 - 122) = 3.488 × 10-17(−316) = -6.54 × 10-18J$

Negative sign is neglected while calculating lambda

$λ = hcΔE = (6.63 × 10-34 J.s)(3.00 × 108m/s)6.54 × 10-18Jλ = 3.04 × 10-8m = 30.4nm ( 1m = 109nm)$

(c)

Interpretation Introduction

Interpretation:

The energy required to eject an electron from $n = 4$ has to be calculated.

Concept Introduction:

Bohr developed a rule for quantization of energy that could be applicable to the electron of an atom in motion. By using this, he derived a formula for energy levels of electron in H-atom.

$E = -RHn2 n = 1,2,3,......(For hydrogen atom)RH is Rydberg constant (2.179 × 10-18 J).Eis energy level.n is principal quantum number.$

(c)

Expert Solution

Answer to Problem 7.119SP

The energy required to eject an electron from $n = 4$ is $2.18 × 10-18J$

Explanation of Solution

The initial state is $n = 4$ and final state is infinity. The energy change for transitions of $4 → ∞$ is calculated as

$ΔE = RHZ2(1ni2 - 1nf2) = (2.18 × 10-18J) (4)2(142 - 1∞2) = (2.18 × 10-18J) (4)2(116 - 0) = (2.18 × 10-18J) (16)(116) = 2.18 × 10-18J$

(d)

Interpretation Introduction

Interpretation:

The physical significance of continuum has to be explained.

(d)

Expert Solution

Explanation of Solution

The energy levels are closely packed when the n values become larger and it leads to continuum of lines. Electrons have been removed from atom when the continuum starts. Therefore, there will be no quantized energy levels associated with electron.

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Answer 2

Question

Chapter 7, Problem 7.119SP

(a)

Interpretation Introduction

Interpretation:

The electronic transition corresponds to line B and C has to be identified.

(a)

Expert Solution

Explanation of Solution

The given lines are corresponding to $n = 2$ . Line A has longest wavelength or lowest energy transition. This indicates the transition is $3 → 2$ transition. Line B has greater wavelength and lower energy than line C. Therefore, line B corresponds to $4 → 2$ and line C corresponds to $5 → 2$ transition.

(b)

Interpretation Introduction

Interpretation:

From the wavelength of line c, the wavelength of line A and B has to be calculated.

Concept Introduction:

Bohr developed a rule for quantization of energy that could be applicable to the electron of an atom in motion. By using this, he derived a formula for energy levels of electron in H-atom.

$E = -RHn2 n = 1,2,3,......(For hydrogen atom)RH is Rydberg constant (2.179 × 10-18 J).Eis energy level.n is principal quantum number.$

(b)

Expert Solution

Answer to Problem 7.119SP

The wavelength of line A is $41.1nm$

The wavelength of line B is $30.4nm$

Explanation of Solution

The energy of line C is calculated using its wavelength as follows,

$E = hcλ = (6.63 × 10-34J.s)(3.00 × 108m/s)(27.1 × 10-9m) = 7.34 × 10-18J$

The atom in which electronic transition of $5 → 2$ occurs is identified as

$ΔE = hν = Ef - EiΔE = RH Z2 (1ni2 - 1nf2)-7.34 × 10-18J = RH Z2 (1ni2 - 1nf2)-7.34 × 10-18J = (2.18 × 10-18J)Z2(152 - 122)-7.34 × 10-18J = Z2(2.18 × 10-18J52 - 2.18 × 10-18J22)-7.34 × 10-18J = Z2(8.72 × 10-18J - 5.45× 10-17J100)-7.34 × 10-18J = (-4.58 × 10-19)Z2Z2 = 16.0Z = 4$

The energy change and wavelength for transitions of $3 → 2$ and $4 → 2$ are calculated as

$ΔE = RHZ2(1ni2 - 1nf2) = (2.18 × 10-18J) (4)2(132 - 122) = (3.488 × 10-17J) (132 - 122) = (3.488 × 10-17J32 - 3.488 × 10-17J22) = (1.395 × 10-16J- 3.139 × 10-16J 36 ) = -4.84 × 10-18J$

Negative sign is neglected while calculating lambda

$λ = hcΔE = (6.63 × 10-34 J.s)(3.00 × 108m/s)4.84 × 10-18Jλ = 4.11 × 10-8m = 41.1nm 1m = 109nm$

$ΔE = RHZ2(1ni2 - 1nf2) = (2.18 × 10-18J) (4)2(142 - 122) = 3.488 × 10-17(−316) = -6.54 × 10-18J$

Negative sign is neglected while calculating lambda

$λ = hcΔE = (6.63 × 10-34 J.s)(3.00 × 108m/s)6.54 × 10-18Jλ = 3.04 × 10-8m = 30.4nm ( 1m = 109nm)$

(c)

Interpretation Introduction

Interpretation:

The energy required to eject an electron from $n = 4$ has to be calculated.

Concept Introduction:

Bohr developed a rule for quantization of energy that could be applicable to the electron of an atom in motion. By using this, he derived a formula for energy levels of electron in H-atom.

$E = -RHn2 n = 1,2,3,......(For hydrogen atom)RH is Rydberg constant (2.179 × 10-18 J).Eis energy level.n is principal quantum number.$

(c)

Expert Solution

Answer to Problem 7.119SP

The energy required to eject an electron from $n = 4$ is $2.18 × 10-18J$

Explanation of Solution

The initial state is $n = 4$ and final state is infinity. The energy change for transitions of $4 → ∞$ is calculated as

$ΔE = RHZ2(1ni2 - 1nf2) = (2.18 × 10-18J) (4)2(142 - 1∞2) = (2.18 × 10-18J) (4)2(116 - 0) = (2.18 × 10-18J) (16)(116) = 2.18 × 10-18J$

(d)

Interpretation Introduction

Interpretation:

The physical significance of continuum has to be explained.

(d)

Expert Solution

Explanation of Solution

The energy levels are closely packed when the n values become larger and it leads to continuum of lines. Electrons have been removed from atom when the continuum starts. Therefore, there will be no quantized energy levels associated with electron.

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Answer 3

Question

Chapter 7, Problem 7.119SP

(a)

Interpretation Introduction

Interpretation:

The electronic transition corresponds to line B and C has to be identified.

(a)

Expert Solution

Explanation of Solution

The given lines are corresponding to $n = 2$ . Line A has longest wavelength or lowest energy transition. This indicates the transition is $3 → 2$ transition. Line B has greater wavelength and lower energy than line C. Therefore, line B corresponds to $4 → 2$ and line C corresponds to $5 → 2$ transition.

(b)

Interpretation Introduction

Interpretation:

From the wavelength of line c, the wavelength of line A and B has to be calculated.

Concept Introduction:

Bohr developed a rule for quantization of energy that could be applicable to the electron of an atom in motion. By using this, he derived a formula for energy levels of electron in H-atom.

$E = -RHn2 n = 1,2,3,......(For hydrogen atom)RH is Rydberg constant (2.179 × 10-18 J).Eis energy level.n is principal quantum number.$

(b)

Expert Solution

Answer to Problem 7.119SP

The wavelength of line A is $41.1nm$

The wavelength of line B is $30.4nm$

Explanation of Solution

The energy of line C is calculated using its wavelength as follows,

$E = hcλ = (6.63 × 10-34J.s)(3.00 × 108m/s)(27.1 × 10-9m) = 7.34 × 10-18J$

The atom in which electronic transition of $5 → 2$ occurs is identified as

$ΔE = hν = Ef - EiΔE = RH Z2 (1ni2 - 1nf2)-7.34 × 10-18J = RH Z2 (1ni2 - 1nf2)-7.34 × 10-18J = (2.18 × 10-18J)Z2(152 - 122)-7.34 × 10-18J = Z2(2.18 × 10-18J52 - 2.18 × 10-18J22)-7.34 × 10-18J = Z2(8.72 × 10-18J - 5.45× 10-17J100)-7.34 × 10-18J = (-4.58 × 10-19)Z2Z2 = 16.0Z = 4$

The energy change and wavelength for transitions of $3 → 2$ and $4 → 2$ are calculated as

$ΔE = RHZ2(1ni2 - 1nf2) = (2.18 × 10-18J) (4)2(132 - 122) = (3.488 × 10-17J) (132 - 122) = (3.488 × 10-17J32 - 3.488 × 10-17J22) = (1.395 × 10-16J- 3.139 × 10-16J 36 ) = -4.84 × 10-18J$

Negative sign is neglected while calculating lambda

$λ = hcΔE = (6.63 × 10-34 J.s)(3.00 × 108m/s)4.84 × 10-18Jλ = 4.11 × 10-8m = 41.1nm 1m = 109nm$

$ΔE = RHZ2(1ni2 - 1nf2) = (2.18 × 10-18J) (4)2(142 - 122) = 3.488 × 10-17(−316) = -6.54 × 10-18J$

Negative sign is neglected while calculating lambda

$λ = hcΔE = (6.63 × 10-34 J.s)(3.00 × 108m/s)6.54 × 10-18Jλ = 3.04 × 10-8m = 30.4nm ( 1m = 109nm)$

(c)

Interpretation Introduction

Interpretation:

The energy required to eject an electron from $n = 4$ has to be calculated.

Concept Introduction:

Bohr developed a rule for quantization of energy that could be applicable to the electron of an atom in motion. By using this, he derived a formula for energy levels of electron in H-atom.

$E = -RHn2 n = 1,2,3,......(For hydrogen atom)RH is Rydberg constant (2.179 × 10-18 J).Eis energy level.n is principal quantum number.$

(c)

Expert Solution

Answer to Problem 7.119SP

The energy required to eject an electron from $n = 4$ is $2.18 × 10-18J$

Explanation of Solution

The initial state is $n = 4$ and final state is infinity. The energy change for transitions of $4 → ∞$ is calculated as

$ΔE = RHZ2(1ni2 - 1nf2) = (2.18 × 10-18J) (4)2(142 - 1∞2) = (2.18 × 10-18J) (4)2(116 - 0) = (2.18 × 10-18J) (16)(116) = 2.18 × 10-18J$

(d)

Interpretation Introduction

Interpretation:

The physical significance of continuum has to be explained.

(d)

Expert Solution

Explanation of Solution

The energy levels are closely packed when the n values become larger and it leads to continuum of lines. Electrons have been removed from atom when the continuum starts. Therefore, there will be no quantized energy levels associated with electron.

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Answer 4

Question

Chapter 7, Problem 7.119SP

(a)

Interpretation Introduction

Interpretation:

The electronic transition corresponds to line B and C has to be identified.

(a)

Expert Solution

Explanation of Solution

The given lines are corresponding to $n = 2$ . Line A has longest wavelength or lowest energy transition. This indicates the transition is $3 → 2$ transition. Line B has greater wavelength and lower energy than line C. Therefore, line B corresponds to $4 → 2$ and line C corresponds to $5 → 2$ transition.

(b)

Interpretation Introduction

Interpretation:

From the wavelength of line c, the wavelength of line A and B has to be calculated.

Concept Introduction:

Bohr developed a rule for quantization of energy that could be applicable to the electron of an atom in motion. By using this, he derived a formula for energy levels of electron in H-atom.

$E = -RHn2 n = 1,2,3,......(For hydrogen atom)RH is Rydberg constant (2.179 × 10-18 J).Eis energy level.n is principal quantum number.$

(b)

Expert Solution

Answer to Problem 7.119SP

The wavelength of line A is $41.1nm$

The wavelength of line B is $30.4nm$

Explanation of Solution

The energy of line C is calculated using its wavelength as follows,

$E = hcλ = (6.63 × 10-34J.s)(3.00 × 108m/s)(27.1 × 10-9m) = 7.34 × 10-18J$

The atom in which electronic transition of $5 → 2$ occurs is identified as

$ΔE = hν = Ef - EiΔE = RH Z2 (1ni2 - 1nf2)-7.34 × 10-18J = RH Z2 (1ni2 - 1nf2)-7.34 × 10-18J = (2.18 × 10-18J)Z2(152 - 122)-7.34 × 10-18J = Z2(2.18 × 10-18J52 - 2.18 × 10-18J22)-7.34 × 10-18J = Z2(8.72 × 10-18J - 5.45× 10-17J100)-7.34 × 10-18J = (-4.58 × 10-19)Z2Z2 = 16.0Z = 4$

The energy change and wavelength for transitions of $3 → 2$ and $4 → 2$ are calculated as

$ΔE = RHZ2(1ni2 - 1nf2) = (2.18 × 10-18J) (4)2(132 - 122) = (3.488 × 10-17J) (132 - 122) = (3.488 × 10-17J32 - 3.488 × 10-17J22) = (1.395 × 10-16J- 3.139 × 10-16J 36 ) = -4.84 × 10-18J$

Negative sign is neglected while calculating lambda

$λ = hcΔE = (6.63 × 10-34 J.s)(3.00 × 108m/s)4.84 × 10-18Jλ = 4.11 × 10-8m = 41.1nm 1m = 109nm$

$ΔE = RHZ2(1ni2 - 1nf2) = (2.18 × 10-18J) (4)2(142 - 122) = 3.488 × 10-17(−316) = -6.54 × 10-18J$

Negative sign is neglected while calculating lambda

$λ = hcΔE = (6.63 × 10-34 J.s)(3.00 × 108m/s)6.54 × 10-18Jλ = 3.04 × 10-8m = 30.4nm ( 1m = 109nm)$

(c)

Interpretation Introduction

Interpretation:

The energy required to eject an electron from $n = 4$ has to be calculated.

Concept Introduction:

Bohr developed a rule for quantization of energy that could be applicable to the electron of an atom in motion. By using this, he derived a formula for energy levels of electron in H-atom.

$E = -RHn2 n = 1,2,3,......(For hydrogen atom)RH is Rydberg constant (2.179 × 10-18 J).Eis energy level.n is principal quantum number.$

(c)

Expert Solution

Answer to Problem 7.119SP

The energy required to eject an electron from $n = 4$ is $2.18 × 10-18J$

Explanation of Solution

The initial state is $n = 4$ and final state is infinity. The energy change for transitions of $4 → ∞$ is calculated as

$ΔE = RHZ2(1ni2 - 1nf2) = (2.18 × 10-18J) (4)2(142 - 1∞2) = (2.18 × 10-18J) (4)2(116 - 0) = (2.18 × 10-18J) (16)(116) = 2.18 × 10-18J$

(d)

Interpretation Introduction

Interpretation:

The physical significance of continuum has to be explained.

(d)

Expert Solution

Explanation of Solution

The energy levels are closely packed when the n values become larger and it leads to continuum of lines. Electrons have been removed from atom when the continuum starts. Therefore, there will be no quantized energy levels associated with electron.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 7, Problem 7.119SP

(a)

Interpretation Introduction

Interpretation:

The electronic transition corresponds to line B and C has to be identified.

(a)

Expert Solution

Explanation of Solution

The given lines are corresponding to $n = 2$ . Line A has longest wavelength or lowest energy transition. This indicates the transition is $3 → 2$ transition. Line B has greater wavelength and lower energy than line C. Therefore, line B corresponds to $4 → 2$ and line C corresponds to $5 → 2$ transition.

(b)

Interpretation Introduction

Interpretation:

From the wavelength of line c, the wavelength of line A and B has to be calculated.

Concept Introduction:

Bohr developed a rule for quantization of energy that could be applicable to the electron of an atom in motion. By using this, he derived a formula for energy levels of electron in H-atom.

$E = -RHn2 n = 1,2,3,......(For hydrogen atom)RH is Rydberg constant (2.179 × 10-18 J).Eis energy level.n is principal quantum number.$

(b)

Expert Solution

Answer to Problem 7.119SP

The wavelength of line A is $41.1nm$

The wavelength of line B is $30.4nm$

Explanation of Solution

The energy of line C is calculated using its wavelength as follows,

$E = hcλ = (6.63 × 10-34J.s)(3.00 × 108m/s)(27.1 × 10-9m) = 7.34 × 10-18J$

The atom in which electronic transition of $5 → 2$ occurs is identified as

$ΔE = hν = Ef - EiΔE = RH Z2 (1ni2 - 1nf2)-7.34 × 10-18J = RH Z2 (1ni2 - 1nf2)-7.34 × 10-18J = (2.18 × 10-18J)Z2(152 - 122)-7.34 × 10-18J = Z2(2.18 × 10-18J52 - 2.18 × 10-18J22)-7.34 × 10-18J = Z2(8.72 × 10-18J - 5.45× 10-17J100)-7.34 × 10-18J = (-4.58 × 10-19)Z2Z2 = 16.0Z = 4$

The energy change and wavelength for transitions of $3 → 2$ and $4 → 2$ are calculated as

$ΔE = RHZ2(1ni2 - 1nf2) = (2.18 × 10-18J) (4)2(132 - 122) = (3.488 × 10-17J) (132 - 122) = (3.488 × 10-17J32 - 3.488 × 10-17J22) = (1.395 × 10-16J- 3.139 × 10-16J 36 ) = -4.84 × 10-18J$

Negative sign is neglected while calculating lambda

$λ = hcΔE = (6.63 × 10-34 J.s)(3.00 × 108m/s)4.84 × 10-18Jλ = 4.11 × 10-8m = 41.1nm 1m = 109nm$

$ΔE = RHZ2(1ni2 - 1nf2) = (2.18 × 10-18J) (4)2(142 - 122) = 3.488 × 10-17(−316) = -6.54 × 10-18J$

Negative sign is neglected while calculating lambda

$λ = hcΔE = (6.63 × 10-34 J.s)(3.00 × 108m/s)6.54 × 10-18Jλ = 3.04 × 10-8m = 30.4nm ( 1m = 109nm)$

(c)

Interpretation Introduction

Interpretation:

The energy required to eject an electron from $n = 4$ has to be calculated.

Concept Introduction:

Bohr developed a rule for quantization of energy that could be applicable to the electron of an atom in motion. By using this, he derived a formula for energy levels of electron in H-atom.

$E = -RHn2 n = 1,2,3,......(For hydrogen atom)RH is Rydberg constant (2.179 × 10-18 J).Eis energy level.n is principal quantum number.$

(c)

Expert Solution

Answer to Problem 7.119SP

The energy required to eject an electron from $n = 4$ is $2.18 × 10-18J$

Explanation of Solution

The initial state is $n = 4$ and final state is infinity. The energy change for transitions of $4 → ∞$ is calculated as

$ΔE = RHZ2(1ni2 - 1nf2) = (2.18 × 10-18J) (4)2(142 - 1∞2) = (2.18 × 10-18J) (4)2(116 - 0) = (2.18 × 10-18J) (16)(116) = 2.18 × 10-18J$

(d)

Interpretation Introduction

Interpretation:

The physical significance of continuum has to be explained.

(d)

Expert Solution

Explanation of Solution

The energy levels are closely packed when the n values become larger and it leads to continuum of lines. Electrons have been removed from atom when the continuum starts. Therefore, there will be no quantized energy levels associated with electron.

Answer 6

Chapter 7, Problem 7.119SP

(a)

Interpretation Introduction

Interpretation:

The electronic transition corresponds to line B and C has to be identified.

(a)

Expert Solution

Explanation of Solution

The given lines are corresponding to $n = 2$ . Line A has longest wavelength or lowest energy transition. This indicates the transition is $3 → 2$ transition. Line B has greater wavelength and lower energy than line C. Therefore, line B corresponds to $4 → 2$ and line C corresponds to $5 → 2$ transition.

Answer 7

Chapter 7, Problem 7.119SP

(a)

Interpretation Introduction

Interpretation:

The electronic transition corresponds to line B and C has to be identified.

Answer 8

(a)

Expert Solution

Explanation of Solution

The given lines are corresponding to $n = 2$ . Line A has longest wavelength or lowest energy transition. This indicates the transition is $3 → 2$ transition. Line B has greater wavelength and lower energy than line C. Therefore, line B corresponds to $4 → 2$ and line C corresponds to $5 → 2$ transition.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

From the wavelength of line c, the wavelength of line A and B has to be calculated.

Concept Introduction:

Bohr developed a rule for quantization of energy that could be applicable to the electron of an atom in motion. By using this, he derived a formula for energy levels of electron in H-atom.

$E = -RHn2 n = 1,2,3,......(For hydrogen atom)RH is Rydberg constant (2.179 × 10-18 J).Eis energy level.n is principal quantum number.$

(b)

Expert Solution

Answer to Problem 7.119SP

The wavelength of line A is $41.1nm$

The wavelength of line B is $30.4nm$

Explanation of Solution

The energy of line C is calculated using its wavelength as follows,

$E = hcλ = (6.63 × 10-34J.s)(3.00 × 108m/s)(27.1 × 10-9m) = 7.34 × 10-18J$

The atom in which electronic transition of $5 → 2$ occurs is identified as

$ΔE = hν = Ef - EiΔE = RH Z2 (1ni2 - 1nf2)-7.34 × 10-18J = RH Z2 (1ni2 - 1nf2)-7.34 × 10-18J = (2.18 × 10-18J)Z2(152 - 122)-7.34 × 10-18J = Z2(2.18 × 10-18J52 - 2.18 × 10-18J22)-7.34 × 10-18J = Z2(8.72 × 10-18J - 5.45× 10-17J100)-7.34 × 10-18J = (-4.58 × 10-19)Z2Z2 = 16.0Z = 4$

The energy change and wavelength for transitions of $3 → 2$ and $4 → 2$ are calculated as

$ΔE = RHZ2(1ni2 - 1nf2) = (2.18 × 10-18J) (4)2(132 - 122) = (3.488 × 10-17J) (132 - 122) = (3.488 × 10-17J32 - 3.488 × 10-17J22) = (1.395 × 10-16J- 3.139 × 10-16J 36 ) = -4.84 × 10-18J$

Negative sign is neglected while calculating lambda

$λ = hcΔE = (6.63 × 10-34 J.s)(3.00 × 108m/s)4.84 × 10-18Jλ = 4.11 × 10-8m = 41.1nm 1m = 109nm$

$ΔE = RHZ2(1ni2 - 1nf2) = (2.18 × 10-18J) (4)2(142 - 122) = 3.488 × 10-17(−316) = -6.54 × 10-18J$

Negative sign is neglected while calculating lambda

$λ = hcΔE = (6.63 × 10-34 J.s)(3.00 × 108m/s)6.54 × 10-18Jλ = 3.04 × 10-8m = 30.4nm ( 1m = 109nm)$

Answer 13

(b)

Interpretation Introduction

Interpretation:

From the wavelength of line c, the wavelength of line A and B has to be calculated.

Concept Introduction:

Bohr developed a rule for quantization of energy that could be applicable to the electron of an atom in motion. By using this, he derived a formula for energy levels of electron in H-atom.

$E = -RHn2 n = 1,2,3,......(For hydrogen atom)RH is Rydberg constant (2.179 × 10-18 J).Eis energy level.n is principal quantum number.$

Answer 14

(b)

Expert Solution

Answer to Problem 7.119SP

The wavelength of line A is $41.1nm$

The wavelength of line B is $30.4nm$

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

Explanation of Solution

The energy of line C is calculated using its wavelength as follows,

$E = hcλ = (6.63 × 10-34J.s)(3.00 × 108m/s)(27.1 × 10-9m) = 7.34 × 10-18J$

The atom in which electronic transition of $5 → 2$ occurs is identified as

$ΔE = hν = Ef - EiΔE = RH Z2 (1ni2 - 1nf2)-7.34 × 10-18J = RH Z2 (1ni2 - 1nf2)-7.34 × 10-18J = (2.18 × 10-18J)Z2(152 - 122)-7.34 × 10-18J = Z2(2.18 × 10-18J52 - 2.18 × 10-18J22)-7.34 × 10-18J = Z2(8.72 × 10-18J - 5.45× 10-17J100)-7.34 × 10-18J = (-4.58 × 10-19)Z2Z2 = 16.0Z = 4$

The energy change and wavelength for transitions of $3 → 2$ and $4 → 2$ are calculated as

$ΔE = RHZ2(1ni2 - 1nf2) = (2.18 × 10-18J) (4)2(132 - 122) = (3.488 × 10-17J) (132 - 122) = (3.488 × 10-17J32 - 3.488 × 10-17J22) = (1.395 × 10-16J- 3.139 × 10-16J 36 ) = -4.84 × 10-18J$

Negative sign is neglected while calculating lambda

$λ = hcΔE = (6.63 × 10-34 J.s)(3.00 × 108m/s)4.84 × 10-18Jλ = 4.11 × 10-8m = 41.1nm 1m = 109nm$

$ΔE = RHZ2(1ni2 - 1nf2) = (2.18 × 10-18J) (4)2(142 - 122) = 3.488 × 10-17(−316) = -6.54 × 10-18J$

Negative sign is neglected while calculating lambda

$λ = hcΔE = (6.63 × 10-34 J.s)(3.00 × 108m/s)6.54 × 10-18Jλ = 3.04 × 10-8m = 30.4nm ( 1m = 109nm)$

Answer 19

(c)

Interpretation Introduction

Interpretation:

The energy required to eject an electron from $n = 4$ has to be calculated.

Concept Introduction:

Bohr developed a rule for quantization of energy that could be applicable to the electron of an atom in motion. By using this, he derived a formula for energy levels of electron in H-atom.

$E = -RHn2 n = 1,2,3,......(For hydrogen atom)RH is Rydberg constant (2.179 × 10-18 J).Eis energy level.n is principal quantum number.$

(c)

Expert Solution

Answer to Problem 7.119SP

The energy required to eject an electron from $n = 4$ is $2.18 × 10-18J$

Explanation of Solution

The initial state is $n = 4$ and final state is infinity. The energy change for transitions of $4 → ∞$ is calculated as

$ΔE = RHZ2(1ni2 - 1nf2) = (2.18 × 10-18J) (4)2(142 - 1∞2) = (2.18 × 10-18J) (4)2(116 - 0) = (2.18 × 10-18J) (16)(116) = 2.18 × 10-18J$

Answer 20

(c)

Interpretation Introduction

Interpretation:

The energy required to eject an electron from $n = 4$ has to be calculated.

Concept Introduction:

Bohr developed a rule for quantization of energy that could be applicable to the electron of an atom in motion. By using this, he derived a formula for energy levels of electron in H-atom.

$E = -RHn2 n = 1,2,3,......(For hydrogen atom)RH is Rydberg constant (2.179 × 10-18 J).Eis energy level.n is principal quantum number.$

Answer 21

(c)

Expert Solution

Answer to Problem 7.119SP

The energy required to eject an electron from $n = 4$ is $2.18 × 10-18J$

Answer 22

(c)

Expert Solution

Answer 23

(c)

Expert Solution

Answer 24

Expert Solution

Answer 25

Explanation of Solution

The initial state is $n = 4$ and final state is infinity. The energy change for transitions of $4 → ∞$ is calculated as

$ΔE = RHZ2(1ni2 - 1nf2) = (2.18 × 10-18J) (4)2(142 - 1∞2) = (2.18 × 10-18J) (4)2(116 - 0) = (2.18 × 10-18J) (16)(116) = 2.18 × 10-18J$

Answer 26

(d)

Interpretation Introduction

Interpretation:

The physical significance of continuum has to be explained.

(d)

Expert Solution

Explanation of Solution

The energy levels are closely packed when the n values become larger and it leads to continuum of lines. Electrons have been removed from atom when the continuum starts. Therefore, there will be no quantized energy levels associated with electron.

Answer 27

(d)

Interpretation Introduction

Interpretation:

The physical significance of continuum has to be explained.

Answer 28

(d)

Expert Solution

Explanation of Solution

The energy levels are closely packed when the n values become larger and it leads to continuum of lines. Electrons have been removed from atom when the continuum starts. Therefore, there will be no quantized energy levels associated with electron.