Energy emitted due to transition.

Question

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Answer 1

Question

Chapter 27, Problem 44E

(a)

To determine

Energy emitted due to transition.

(a)

Expert Solution

Answer to Problem 44E

The photon of energy emitted is 1081 eV.

Explanation of Solution

Write the expression for energy of diatomic molecule.

En=n(n+1)ℏ22I (I)

Here, En is the energy of nth level, n is the number of level, ℏ is the reduced Planck’s constant and I is the moment of inertia.

Write the expression for change in energy due to transition.

ΔE=E2−E0 (II)

Conclusion:

Substitute 2 for n, 1.055×10−34 Js for ℏ and 1.92×10−46 kgm2 for I in equation (I).

E2=2(2+1)(1.055×10−34 Js)22(1.92×10−46 kgm2)=2(3)(1.055×10−34 Js)22(1.92×10−46 kgm2)(1 kgm2s−21 J)=1.73×10−22 J(1 eV1.6×10−19 J)=1081 eV

Substitute 0 for n, 1.055×10−34 Js for ℏ and 1.92×10−46 kgm2 for I in equation (I).

E0=(0)(0+1)(1.055×10−34 Js)22(1.92×10−46 kgm2)=0

Substitute 1081 eV for E2 and 0 eV for E0 in equation (II)>

ΔE=1081 eV−0 eV=1081 eV

Thus, the energy of photon emitted is 1081 eV.

(b)

To determine

The wavelength of photon emitted.

(b)

Expert Solution

Answer to Problem 44E

The wavelength of photon emitted is 11.49 nm.

Explanation of Solution

Write the expression for energy of photon emitted.

E=hcλ

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

Rearrange the above expression for λ.

λ=hcE (III)

Conclusion:

Substitute 6.626×10−34 Js for h, 3×108 ms−1 for c and 1081 eV for E in equation (III).

λ=(6.626×10−34 Js)(3×108ms−1)1081 eV=(6.626×10−34 Js)(3×108ms−1)1081 eV(1 kgm2s−21 J)(1 eV1.6×10−19 J)=1.149×10−8 m(1 nm10−9 m)=11.49 nm

Thus, the wavelength of photon emitted is 11.49 nm.

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Answer 2

Question

Chapter 27, Problem 44E

(a)

To determine

Energy emitted due to transition.

(a)

Expert Solution

Answer to Problem 44E

The photon of energy emitted is 1081 eV.

Explanation of Solution

Write the expression for energy of diatomic molecule.

En=n(n+1)ℏ22I (I)

Here, En is the energy of nth level, n is the number of level, ℏ is the reduced Planck’s constant and I is the moment of inertia.

Write the expression for change in energy due to transition.

ΔE=E2−E0 (II)

Conclusion:

Substitute 2 for n, 1.055×10−34 Js for ℏ and 1.92×10−46 kgm2 for I in equation (I).

E2=2(2+1)(1.055×10−34 Js)22(1.92×10−46 kgm2)=2(3)(1.055×10−34 Js)22(1.92×10−46 kgm2)(1 kgm2s−21 J)=1.73×10−22 J(1 eV1.6×10−19 J)=1081 eV

Substitute 0 for n, 1.055×10−34 Js for ℏ and 1.92×10−46 kgm2 for I in equation (I).

E0=(0)(0+1)(1.055×10−34 Js)22(1.92×10−46 kgm2)=0

Substitute 1081 eV for E2 and 0 eV for E0 in equation (II)>

ΔE=1081 eV−0 eV=1081 eV

Thus, the energy of photon emitted is 1081 eV.

(b)

To determine

The wavelength of photon emitted.

(b)

Expert Solution

Answer to Problem 44E

The wavelength of photon emitted is 11.49 nm.

Explanation of Solution

Write the expression for energy of photon emitted.

E=hcλ

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

Rearrange the above expression for λ.

λ=hcE (III)

Conclusion:

Substitute 6.626×10−34 Js for h, 3×108 ms−1 for c and 1081 eV for E in equation (III).

λ=(6.626×10−34 Js)(3×108ms−1)1081 eV=(6.626×10−34 Js)(3×108ms−1)1081 eV(1 kgm2s−21 J)(1 eV1.6×10−19 J)=1.149×10−8 m(1 nm10−9 m)=11.49 nm

Thus, the wavelength of photon emitted is 11.49 nm.

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Answer 3

Question

Chapter 27, Problem 44E

(a)

To determine

Energy emitted due to transition.

(a)

Expert Solution

Answer to Problem 44E

The photon of energy emitted is 1081 eV.

Explanation of Solution

Write the expression for energy of diatomic molecule.

En=n(n+1)ℏ22I (I)

Here, En is the energy of nth level, n is the number of level, ℏ is the reduced Planck’s constant and I is the moment of inertia.

Write the expression for change in energy due to transition.

ΔE=E2−E0 (II)

Conclusion:

Substitute 2 for n, 1.055×10−34 Js for ℏ and 1.92×10−46 kgm2 for I in equation (I).

E2=2(2+1)(1.055×10−34 Js)22(1.92×10−46 kgm2)=2(3)(1.055×10−34 Js)22(1.92×10−46 kgm2)(1 kgm2s−21 J)=1.73×10−22 J(1 eV1.6×10−19 J)=1081 eV

Substitute 0 for n, 1.055×10−34 Js for ℏ and 1.92×10−46 kgm2 for I in equation (I).

E0=(0)(0+1)(1.055×10−34 Js)22(1.92×10−46 kgm2)=0

Substitute 1081 eV for E2 and 0 eV for E0 in equation (II)>

ΔE=1081 eV−0 eV=1081 eV

Thus, the energy of photon emitted is 1081 eV.

(b)

To determine

The wavelength of photon emitted.

(b)

Expert Solution

Answer to Problem 44E

The wavelength of photon emitted is 11.49 nm.

Explanation of Solution

Write the expression for energy of photon emitted.

E=hcλ

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

Rearrange the above expression for λ.

λ=hcE (III)

Conclusion:

Substitute 6.626×10−34 Js for h, 3×108 ms−1 for c and 1081 eV for E in equation (III).

λ=(6.626×10−34 Js)(3×108ms−1)1081 eV=(6.626×10−34 Js)(3×108ms−1)1081 eV(1 kgm2s−21 J)(1 eV1.6×10−19 J)=1.149×10−8 m(1 nm10−9 m)=11.49 nm

Thus, the wavelength of photon emitted is 11.49 nm.

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Answer 4

Question

Chapter 27, Problem 44E

(a)

To determine

Energy emitted due to transition.

(a)

Expert Solution

Answer to Problem 44E

The photon of energy emitted is 1081 eV.

Explanation of Solution

Write the expression for energy of diatomic molecule.

En=n(n+1)ℏ22I (I)

Here, En is the energy of nth level, n is the number of level, ℏ is the reduced Planck’s constant and I is the moment of inertia.

Write the expression for change in energy due to transition.

ΔE=E2−E0 (II)

Conclusion:

Substitute 2 for n, 1.055×10−34 Js for ℏ and 1.92×10−46 kgm2 for I in equation (I).

E2=2(2+1)(1.055×10−34 Js)22(1.92×10−46 kgm2)=2(3)(1.055×10−34 Js)22(1.92×10−46 kgm2)(1 kgm2s−21 J)=1.73×10−22 J(1 eV1.6×10−19 J)=1081 eV

Substitute 0 for n, 1.055×10−34 Js for ℏ and 1.92×10−46 kgm2 for I in equation (I).

E0=(0)(0+1)(1.055×10−34 Js)22(1.92×10−46 kgm2)=0

Substitute 1081 eV for E2 and 0 eV for E0 in equation (II)>

ΔE=1081 eV−0 eV=1081 eV

Thus, the energy of photon emitted is 1081 eV.

(b)

To determine

The wavelength of photon emitted.

(b)

Expert Solution

Answer to Problem 44E

The wavelength of photon emitted is 11.49 nm.

Explanation of Solution

Write the expression for energy of photon emitted.

E=hcλ

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

Rearrange the above expression for λ.

λ=hcE (III)

Conclusion:

Substitute 6.626×10−34 Js for h, 3×108 ms−1 for c and 1081 eV for E in equation (III).

λ=(6.626×10−34 Js)(3×108ms−1)1081 eV=(6.626×10−34 Js)(3×108ms−1)1081 eV(1 kgm2s−21 J)(1 eV1.6×10−19 J)=1.149×10−8 m(1 nm10−9 m)=11.49 nm

Thus, the wavelength of photon emitted is 11.49 nm.

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Answer 5

Chapter 27, Problem 44E

(a)

To determine

Energy emitted due to transition.

(a)

Expert Solution

Answer to Problem 44E

The photon of energy emitted is 1081 eV.

Explanation of Solution

Write the expression for energy of diatomic molecule.

En=n(n+1)ℏ22I (I)

Here, En is the energy of nth level, n is the number of level, ℏ is the reduced Planck’s constant and I is the moment of inertia.

Write the expression for change in energy due to transition.

ΔE=E2−E0 (II)

Conclusion:

Substitute 2 for n, 1.055×10−34 Js for ℏ and 1.92×10−46 kgm2 for I in equation (I).

E2=2(2+1)(1.055×10−34 Js)22(1.92×10−46 kgm2)=2(3)(1.055×10−34 Js)22(1.92×10−46 kgm2)(1 kgm2s−21 J)=1.73×10−22 J(1 eV1.6×10−19 J)=1081 eV

Substitute 0 for n, 1.055×10−34 Js for ℏ and 1.92×10−46 kgm2 for I in equation (I).

E0=(0)(0+1)(1.055×10−34 Js)22(1.92×10−46 kgm2)=0

Substitute 1081 eV for E2 and 0 eV for E0 in equation (II)>

ΔE=1081 eV−0 eV=1081 eV

Thus, the energy of photon emitted is 1081 eV.

(b)

To determine

The wavelength of photon emitted.

(b)

Expert Solution

Answer to Problem 44E

The wavelength of photon emitted is 11.49 nm.

Explanation of Solution

Write the expression for energy of photon emitted.

E=hcλ

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

Rearrange the above expression for λ.

λ=hcE (III)

Conclusion:

Substitute 6.626×10−34 Js for h, 3×108 ms−1 for c and 1081 eV for E in equation (III).

λ=(6.626×10−34 Js)(3×108ms−1)1081 eV=(6.626×10−34 Js)(3×108ms−1)1081 eV(1 kgm2s−21 J)(1 eV1.6×10−19 J)=1.149×10−8 m(1 nm10−9 m)=11.49 nm

Thus, the wavelength of photon emitted is 11.49 nm.

Answer 6

Chapter 27, Problem 44E

(a)

To determine

Energy emitted due to transition.

(a)

Expert Solution

Answer to Problem 44E

The photon of energy emitted is 1081 eV.

Explanation of Solution

Write the expression for energy of diatomic molecule.

En=n(n+1)ℏ22I (I)

Here, En is the energy of nth level, n is the number of level, ℏ is the reduced Planck’s constant and I is the moment of inertia.

Write the expression for change in energy due to transition.

ΔE=E2−E0 (II)

Conclusion:

Substitute 2 for n, 1.055×10−34 Js for ℏ and 1.92×10−46 kgm2 for I in equation (I).

E2=2(2+1)(1.055×10−34 Js)22(1.92×10−46 kgm2)=2(3)(1.055×10−34 Js)22(1.92×10−46 kgm2)(1 kgm2s−21 J)=1.73×10−22 J(1 eV1.6×10−19 J)=1081 eV

Substitute 0 for n, 1.055×10−34 Js for ℏ and 1.92×10−46 kgm2 for I in equation (I).

E0=(0)(0+1)(1.055×10−34 Js)22(1.92×10−46 kgm2)=0

Substitute 1081 eV for E2 and 0 eV for E0 in equation (II)>

ΔE=1081 eV−0 eV=1081 eV

Thus, the energy of photon emitted is 1081 eV.

Answer 7

Chapter 27, Problem 44E

(a)

To determine

Energy emitted due to transition.

Answer 8

(a)

Expert Solution

Answer to Problem 44E

The photon of energy emitted is 1081 eV.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write the expression for energy of diatomic molecule.

En=n(n+1)ℏ22I (I)

Here, En is the energy of nth level, n is the number of level, ℏ is the reduced Planck’s constant and I is the moment of inertia.

Write the expression for change in energy due to transition.

ΔE=E2−E0 (II)

Conclusion:

Substitute 2 for n, 1.055×10−34 Js for ℏ and 1.92×10−46 kgm2 for I in equation (I).

E2=2(2+1)(1.055×10−34 Js)22(1.92×10−46 kgm2)=2(3)(1.055×10−34 Js)22(1.92×10−46 kgm2)(1 kgm2s−21 J)=1.73×10−22 J(1 eV1.6×10−19 J)=1081 eV

Substitute 0 for n, 1.055×10−34 Js for ℏ and 1.92×10−46 kgm2 for I in equation (I).

E0=(0)(0+1)(1.055×10−34 Js)22(1.92×10−46 kgm2)=0

Substitute 1081 eV for E2 and 0 eV for E0 in equation (II)>

ΔE=1081 eV−0 eV=1081 eV

Thus, the energy of photon emitted is 1081 eV.

Answer 13

(b)

To determine

The wavelength of photon emitted.

(b)

Expert Solution

Answer to Problem 44E

The wavelength of photon emitted is 11.49 nm.

Explanation of Solution

Write the expression for energy of photon emitted.

E=hcλ

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

Rearrange the above expression for λ.

λ=hcE (III)

Conclusion:

Substitute 6.626×10−34 Js for h, 3×108 ms−1 for c and 1081 eV for E in equation (III).

λ=(6.626×10−34 Js)(3×108ms−1)1081 eV=(6.626×10−34 Js)(3×108ms−1)1081 eV(1 kgm2s−21 J)(1 eV1.6×10−19 J)=1.149×10−8 m(1 nm10−9 m)=11.49 nm

Thus, the wavelength of photon emitted is 11.49 nm.

Answer 14

(b)

To determine

The wavelength of photon emitted.

Answer 15

(b)

Expert Solution

Answer to Problem 44E

The wavelength of photon emitted is 11.49 nm.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Write the expression for energy of photon emitted.

E=hcλ

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

Rearrange the above expression for λ.

λ=hcE (III)

Conclusion:

Substitute 6.626×10−34 Js for h, 3×108 ms−1 for c and 1081 eV for E in equation (III).

λ=(6.626×10−34 Js)(3×108ms−1)1081 eV=(6.626×10−34 Js)(3×108ms−1)1081 eV(1 kgm2s−21 J)(1 eV1.6×10−19 J)=1.149×10−8 m(1 nm10−9 m)=11.49 nm

Thus, the wavelength of photon emitted is 11.49 nm.

Answer 20

Ch. 27 - Prob. 1RQ Ch. 27 - Prob. 2RQ Ch. 27 - Prob. 3RQ Ch. 27 - Prob. 4RQ Ch. 27 - Prob. 5RQ Ch. 27 - Prob. 6RQ Ch. 27 - Prob. 7RQ Ch. 27 - Prob. 8RQ Ch. 27 - Prob. 9RQ Ch. 27 - Prob. 10RQ

Energy emitted due to transition.

Energy emitted due to transition.

Answer to Problem 44E

Explanation of Solution

The wavelength of photon emitted.

Answer to Problem 44E

Explanation of Solution

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Chapter 27 Solutions