Frequency, wavenumber and energy ( J/photon and kJ/mol ) of photon with 250 nm have to be calculated. Concept Introduction: Planck’s hypothesis states energy associated with photon is as follows: E = h c λ Here, λ denotes wavelength. h denotes Planck’s constant c denotes speed of light.

Question

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Answer 1

Question

Chapter 18, Problem 18.9P

(a)

Interpretation Introduction

Interpretation:

Frequency, wavenumber and energy (J/photon and kJ/mol) of photon with 250 nm have to be calculated.

Concept Introduction:

Planck’s hypothesis states energy associated with photon is as follows:

E=hcλ

Here,

λ denotes wavelength.

h denotes Planck’s constant

c denotes speed of light.

(a)

Expert Solution

Explanation of Solution

Conversion factor to convert nm to m is as follows:

1 nm=10−9 m

Thus, 250 nm is converted to m as follows:

Wavelength=(250 nm )(10−9 m1 nm)=2.5×10−7 m

Formula to calculate frequency is as follows:

v=cλ (1)

Substitute 2.5×10−7 m for λ , and 3×108 m/s for c in equation (1).

v=3×108 m/s2.5×10−7 m=1.2×1015 Hz

Formula to calculate wavenumber is as follows:

v¯=vc (2)

Substitute 1.2×1015 Hz for v , and 3×108 m/s for c in equation (2).

v¯=(1.2×1015 Hz3×108 m/s)(1 m102 cm)=4.0×104 cm−1

Planck’s hypothesis states energy associated with single photon is as follows:

E=hcλ (3)

Substitute 2.5×10−7 m for λ , 6.626×10−34 J s for h and 3×108 m/s for c in equation (3).

E=(6.626×10−34 J s)(3×108 m/s)(2.5×10−7 m)=7.95×10−19 J/photon

Avogadro’s number is 6.022×1023 photons/mol. So energy for 1 mol photons is calculated as follows:

E=(6.022×1023 photons/mol)(7.95×10−19 J1 photon)(1 kJ1000 J)=478.7 kJ/mol≈479 kJ/mol

Thus, frequency, wavenumber and energy (J/photon and kJ/mol) are 1.2×1015 Hz, 4.0×103 cm−1, 7.95×10−19 J/photon and 479 kJ/mol respectively.

(b)

Interpretation Introduction

Interpretation:

Frequency, wavenumber and energy (J/photon and kJ/mol) of photon with 2.50 μm have to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Conversion factor to convert μm to m is as follows:

1 μm=10−6 m

Thus, 2.50 μm is converted to m as follows:

Wavelength=(2.50 μm )(10−6 m1 μm)=2.5×10−6 m

Substitute 2.5×10−6 m for λ , and 3×108 m/s for c in equation (1).

v=3×108 m/s2.5×10−6 m=1.2×1014 Hz

Substitute 1.2×1014 Hz for v , and 3×108 m/s for c in equation (2).

v¯=(1.2×1014 Hz3×108 m/s)(1 m102 cm)=4.0×103 cm−1

Substitute 2.5×10−6 m for λ , 6.626×10−34 J s for h and 3×108 m/s for c in equation (3).

E=(6.626×10−34 J s)(3×108 m/s)(2.5×10−6 m)=7.95×10−20 J/photon

Avogadro’s number is 6.022×1023 photons/mol. So energy for 1 mol photons is calculated as follows:

E=(6.022×1023 photons/mol)(7.95×10−20 J1 photon)(1 kJ1000 J)=47.87 kJ/mol≈47.9 kJ/mol

Thus, frequency, wavenumber and energy (J/photon and kJ/mol) are 1.2×1014 Hz, 4.0×103 cm−1, 7.95×10−20 J/photon and 47.9 kJ/mol respectively.

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Answer 2

Question

Chapter 18, Problem 18.9P

(a)

Interpretation Introduction

Interpretation:

Frequency, wavenumber and energy (J/photon and kJ/mol) of photon with 250 nm have to be calculated.

Concept Introduction:

Planck’s hypothesis states energy associated with photon is as follows:

E=hcλ

Here,

λ denotes wavelength.

h denotes Planck’s constant

c denotes speed of light.

(a)

Expert Solution

Explanation of Solution

Conversion factor to convert nm to m is as follows:

1 nm=10−9 m

Thus, 250 nm is converted to m as follows:

Wavelength=(250 nm )(10−9 m1 nm)=2.5×10−7 m

Formula to calculate frequency is as follows:

v=cλ (1)

Substitute 2.5×10−7 m for λ , and 3×108 m/s for c in equation (1).

v=3×108 m/s2.5×10−7 m=1.2×1015 Hz

Formula to calculate wavenumber is as follows:

v¯=vc (2)

Substitute 1.2×1015 Hz for v , and 3×108 m/s for c in equation (2).

v¯=(1.2×1015 Hz3×108 m/s)(1 m102 cm)=4.0×104 cm−1

Planck’s hypothesis states energy associated with single photon is as follows:

E=hcλ (3)

Substitute 2.5×10−7 m for λ , 6.626×10−34 J s for h and 3×108 m/s for c in equation (3).

E=(6.626×10−34 J s)(3×108 m/s)(2.5×10−7 m)=7.95×10−19 J/photon

Avogadro’s number is 6.022×1023 photons/mol. So energy for 1 mol photons is calculated as follows:

E=(6.022×1023 photons/mol)(7.95×10−19 J1 photon)(1 kJ1000 J)=478.7 kJ/mol≈479 kJ/mol

Thus, frequency, wavenumber and energy (J/photon and kJ/mol) are 1.2×1015 Hz, 4.0×103 cm−1, 7.95×10−19 J/photon and 479 kJ/mol respectively.

(b)

Interpretation Introduction

Interpretation:

Frequency, wavenumber and energy (J/photon and kJ/mol) of photon with 2.50 μm have to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Conversion factor to convert μm to m is as follows:

1 μm=10−6 m

Thus, 2.50 μm is converted to m as follows:

Wavelength=(2.50 μm )(10−6 m1 μm)=2.5×10−6 m

Substitute 2.5×10−6 m for λ , and 3×108 m/s for c in equation (1).

v=3×108 m/s2.5×10−6 m=1.2×1014 Hz

Substitute 1.2×1014 Hz for v , and 3×108 m/s for c in equation (2).

v¯=(1.2×1014 Hz3×108 m/s)(1 m102 cm)=4.0×103 cm−1

Substitute 2.5×10−6 m for λ , 6.626×10−34 J s for h and 3×108 m/s for c in equation (3).

E=(6.626×10−34 J s)(3×108 m/s)(2.5×10−6 m)=7.95×10−20 J/photon

Avogadro’s number is 6.022×1023 photons/mol. So energy for 1 mol photons is calculated as follows:

E=(6.022×1023 photons/mol)(7.95×10−20 J1 photon)(1 kJ1000 J)=47.87 kJ/mol≈47.9 kJ/mol

Thus, frequency, wavenumber and energy (J/photon and kJ/mol) are 1.2×1014 Hz, 4.0×103 cm−1, 7.95×10−20 J/photon and 47.9 kJ/mol respectively.

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Answer 3

Question

Chapter 18, Problem 18.9P

(a)

Interpretation Introduction

Interpretation:

Frequency, wavenumber and energy (J/photon and kJ/mol) of photon with 250 nm have to be calculated.

Concept Introduction:

Planck’s hypothesis states energy associated with photon is as follows:

E=hcλ

Here,

λ denotes wavelength.

h denotes Planck’s constant

c denotes speed of light.

(a)

Expert Solution

Explanation of Solution

Conversion factor to convert nm to m is as follows:

1 nm=10−9 m

Thus, 250 nm is converted to m as follows:

Wavelength=(250 nm )(10−9 m1 nm)=2.5×10−7 m

Formula to calculate frequency is as follows:

v=cλ (1)

Substitute 2.5×10−7 m for λ , and 3×108 m/s for c in equation (1).

v=3×108 m/s2.5×10−7 m=1.2×1015 Hz

Formula to calculate wavenumber is as follows:

v¯=vc (2)

Substitute 1.2×1015 Hz for v , and 3×108 m/s for c in equation (2).

v¯=(1.2×1015 Hz3×108 m/s)(1 m102 cm)=4.0×104 cm−1

Planck’s hypothesis states energy associated with single photon is as follows:

E=hcλ (3)

Substitute 2.5×10−7 m for λ , 6.626×10−34 J s for h and 3×108 m/s for c in equation (3).

E=(6.626×10−34 J s)(3×108 m/s)(2.5×10−7 m)=7.95×10−19 J/photon

Avogadro’s number is 6.022×1023 photons/mol. So energy for 1 mol photons is calculated as follows:

E=(6.022×1023 photons/mol)(7.95×10−19 J1 photon)(1 kJ1000 J)=478.7 kJ/mol≈479 kJ/mol

Thus, frequency, wavenumber and energy (J/photon and kJ/mol) are 1.2×1015 Hz, 4.0×103 cm−1, 7.95×10−19 J/photon and 479 kJ/mol respectively.

(b)

Interpretation Introduction

Interpretation:

Frequency, wavenumber and energy (J/photon and kJ/mol) of photon with 2.50 μm have to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Conversion factor to convert μm to m is as follows:

1 μm=10−6 m

Thus, 2.50 μm is converted to m as follows:

Wavelength=(2.50 μm )(10−6 m1 μm)=2.5×10−6 m

Substitute 2.5×10−6 m for λ , and 3×108 m/s for c in equation (1).

v=3×108 m/s2.5×10−6 m=1.2×1014 Hz

Substitute 1.2×1014 Hz for v , and 3×108 m/s for c in equation (2).

v¯=(1.2×1014 Hz3×108 m/s)(1 m102 cm)=4.0×103 cm−1

Substitute 2.5×10−6 m for λ , 6.626×10−34 J s for h and 3×108 m/s for c in equation (3).

E=(6.626×10−34 J s)(3×108 m/s)(2.5×10−6 m)=7.95×10−20 J/photon

Avogadro’s number is 6.022×1023 photons/mol. So energy for 1 mol photons is calculated as follows:

E=(6.022×1023 photons/mol)(7.95×10−20 J1 photon)(1 kJ1000 J)=47.87 kJ/mol≈47.9 kJ/mol

Thus, frequency, wavenumber and energy (J/photon and kJ/mol) are 1.2×1014 Hz, 4.0×103 cm−1, 7.95×10−20 J/photon and 47.9 kJ/mol respectively.

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Answer 4

Question

Chapter 18, Problem 18.9P

(a)

Interpretation Introduction

Interpretation:

Frequency, wavenumber and energy (J/photon and kJ/mol) of photon with 250 nm have to be calculated.

Concept Introduction:

Planck’s hypothesis states energy associated with photon is as follows:

E=hcλ

Here,

λ denotes wavelength.

h denotes Planck’s constant

c denotes speed of light.

(a)

Expert Solution

Explanation of Solution

Conversion factor to convert nm to m is as follows:

1 nm=10−9 m

Thus, 250 nm is converted to m as follows:

Wavelength=(250 nm )(10−9 m1 nm)=2.5×10−7 m

Formula to calculate frequency is as follows:

v=cλ (1)

Substitute 2.5×10−7 m for λ , and 3×108 m/s for c in equation (1).

v=3×108 m/s2.5×10−7 m=1.2×1015 Hz

Formula to calculate wavenumber is as follows:

v¯=vc (2)

Substitute 1.2×1015 Hz for v , and 3×108 m/s for c in equation (2).

v¯=(1.2×1015 Hz3×108 m/s)(1 m102 cm)=4.0×104 cm−1

Planck’s hypothesis states energy associated with single photon is as follows:

E=hcλ (3)

Substitute 2.5×10−7 m for λ , 6.626×10−34 J s for h and 3×108 m/s for c in equation (3).

E=(6.626×10−34 J s)(3×108 m/s)(2.5×10−7 m)=7.95×10−19 J/photon

Avogadro’s number is 6.022×1023 photons/mol. So energy for 1 mol photons is calculated as follows:

E=(6.022×1023 photons/mol)(7.95×10−19 J1 photon)(1 kJ1000 J)=478.7 kJ/mol≈479 kJ/mol

Thus, frequency, wavenumber and energy (J/photon and kJ/mol) are 1.2×1015 Hz, 4.0×103 cm−1, 7.95×10−19 J/photon and 479 kJ/mol respectively.

(b)

Interpretation Introduction

Interpretation:

Frequency, wavenumber and energy (J/photon and kJ/mol) of photon with 2.50 μm have to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Conversion factor to convert μm to m is as follows:

1 μm=10−6 m

Thus, 2.50 μm is converted to m as follows:

Wavelength=(2.50 μm )(10−6 m1 μm)=2.5×10−6 m

Substitute 2.5×10−6 m for λ , and 3×108 m/s for c in equation (1).

v=3×108 m/s2.5×10−6 m=1.2×1014 Hz

Substitute 1.2×1014 Hz for v , and 3×108 m/s for c in equation (2).

v¯=(1.2×1014 Hz3×108 m/s)(1 m102 cm)=4.0×103 cm−1

Substitute 2.5×10−6 m for λ , 6.626×10−34 J s for h and 3×108 m/s for c in equation (3).

E=(6.626×10−34 J s)(3×108 m/s)(2.5×10−6 m)=7.95×10−20 J/photon

Avogadro’s number is 6.022×1023 photons/mol. So energy for 1 mol photons is calculated as follows:

E=(6.022×1023 photons/mol)(7.95×10−20 J1 photon)(1 kJ1000 J)=47.87 kJ/mol≈47.9 kJ/mol

Thus, frequency, wavenumber and energy (J/photon and kJ/mol) are 1.2×1014 Hz, 4.0×103 cm−1, 7.95×10−20 J/photon and 47.9 kJ/mol respectively.

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Answer 5

Chapter 18, Problem 18.9P

(a)

Interpretation Introduction

Interpretation:

Frequency, wavenumber and energy (J/photon and kJ/mol) of photon with 250 nm have to be calculated.

Concept Introduction:

Planck’s hypothesis states energy associated with photon is as follows:

E=hcλ

Here,

λ denotes wavelength.

h denotes Planck’s constant

c denotes speed of light.

(a)

Expert Solution

Explanation of Solution

Conversion factor to convert nm to m is as follows:

1 nm=10−9 m

Thus, 250 nm is converted to m as follows:

Wavelength=(250 nm )(10−9 m1 nm)=2.5×10−7 m

Formula to calculate frequency is as follows:

v=cλ (1)

Substitute 2.5×10−7 m for λ , and 3×108 m/s for c in equation (1).

v=3×108 m/s2.5×10−7 m=1.2×1015 Hz

Formula to calculate wavenumber is as follows:

v¯=vc (2)

Substitute 1.2×1015 Hz for v , and 3×108 m/s for c in equation (2).

v¯=(1.2×1015 Hz3×108 m/s)(1 m102 cm)=4.0×104 cm−1

Planck’s hypothesis states energy associated with single photon is as follows:

E=hcλ (3)

Substitute 2.5×10−7 m for λ , 6.626×10−34 J s for h and 3×108 m/s for c in equation (3).

E=(6.626×10−34 J s)(3×108 m/s)(2.5×10−7 m)=7.95×10−19 J/photon

Avogadro’s number is 6.022×1023 photons/mol. So energy for 1 mol photons is calculated as follows:

E=(6.022×1023 photons/mol)(7.95×10−19 J1 photon)(1 kJ1000 J)=478.7 kJ/mol≈479 kJ/mol

Thus, frequency, wavenumber and energy (J/photon and kJ/mol) are 1.2×1015 Hz, 4.0×103 cm−1, 7.95×10−19 J/photon and 479 kJ/mol respectively.

(b)

Interpretation Introduction

Interpretation:

Frequency, wavenumber and energy (J/photon and kJ/mol) of photon with 2.50 μm have to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Conversion factor to convert μm to m is as follows:

1 μm=10−6 m

Thus, 2.50 μm is converted to m as follows:

Wavelength=(2.50 μm )(10−6 m1 μm)=2.5×10−6 m

Substitute 2.5×10−6 m for λ , and 3×108 m/s for c in equation (1).

v=3×108 m/s2.5×10−6 m=1.2×1014 Hz

Substitute 1.2×1014 Hz for v , and 3×108 m/s for c in equation (2).

v¯=(1.2×1014 Hz3×108 m/s)(1 m102 cm)=4.0×103 cm−1

Substitute 2.5×10−6 m for λ , 6.626×10−34 J s for h and 3×108 m/s for c in equation (3).

E=(6.626×10−34 J s)(3×108 m/s)(2.5×10−6 m)=7.95×10−20 J/photon

Avogadro’s number is 6.022×1023 photons/mol. So energy for 1 mol photons is calculated as follows:

E=(6.022×1023 photons/mol)(7.95×10−20 J1 photon)(1 kJ1000 J)=47.87 kJ/mol≈47.9 kJ/mol

Thus, frequency, wavenumber and energy (J/photon and kJ/mol) are 1.2×1014 Hz, 4.0×103 cm−1, 7.95×10−20 J/photon and 47.9 kJ/mol respectively.

Answer 6

Chapter 18, Problem 18.9P

(a)

Interpretation Introduction

Interpretation:

Frequency, wavenumber and energy (J/photon and kJ/mol) of photon with 250 nm have to be calculated.

Concept Introduction:

Planck’s hypothesis states energy associated with photon is as follows:

E=hcλ

Here,

λ denotes wavelength.

h denotes Planck’s constant

c denotes speed of light.

(a)

Expert Solution

Explanation of Solution

Conversion factor to convert nm to m is as follows:

1 nm=10−9 m

Thus, 250 nm is converted to m as follows:

Wavelength=(250 nm )(10−9 m1 nm)=2.5×10−7 m

Formula to calculate frequency is as follows:

v=cλ (1)

Substitute 2.5×10−7 m for λ , and 3×108 m/s for c in equation (1).

v=3×108 m/s2.5×10−7 m=1.2×1015 Hz

Formula to calculate wavenumber is as follows:

v¯=vc (2)

Substitute 1.2×1015 Hz for v , and 3×108 m/s for c in equation (2).

v¯=(1.2×1015 Hz3×108 m/s)(1 m102 cm)=4.0×104 cm−1

Planck’s hypothesis states energy associated with single photon is as follows:

E=hcλ (3)

Substitute 2.5×10−7 m for λ , 6.626×10−34 J s for h and 3×108 m/s for c in equation (3).

E=(6.626×10−34 J s)(3×108 m/s)(2.5×10−7 m)=7.95×10−19 J/photon

Avogadro’s number is 6.022×1023 photons/mol. So energy for 1 mol photons is calculated as follows:

E=(6.022×1023 photons/mol)(7.95×10−19 J1 photon)(1 kJ1000 J)=478.7 kJ/mol≈479 kJ/mol

Thus, frequency, wavenumber and energy (J/photon and kJ/mol) are 1.2×1015 Hz, 4.0×103 cm−1, 7.95×10−19 J/photon and 479 kJ/mol respectively.

Answer 7

Chapter 18, Problem 18.9P

(a)

Interpretation Introduction

Interpretation:

Frequency, wavenumber and energy (J/photon and kJ/mol) of photon with 250 nm have to be calculated.

Concept Introduction:

Planck’s hypothesis states energy associated with photon is as follows:

E=hcλ

Here,

λ denotes wavelength.

h denotes Planck’s constant

c denotes speed of light.

Answer 8

(a)

Expert Solution

Explanation of Solution

Conversion factor to convert nm to m is as follows:

1 nm=10−9 m

Thus, 250 nm is converted to m as follows:

Wavelength=(250 nm )(10−9 m1 nm)=2.5×10−7 m

Formula to calculate frequency is as follows:

v=cλ (1)

Substitute 2.5×10−7 m for λ , and 3×108 m/s for c in equation (1).

v=3×108 m/s2.5×10−7 m=1.2×1015 Hz

Formula to calculate wavenumber is as follows:

v¯=vc (2)

Substitute 1.2×1015 Hz for v , and 3×108 m/s for c in equation (2).

v¯=(1.2×1015 Hz3×108 m/s)(1 m102 cm)=4.0×104 cm−1

Planck’s hypothesis states energy associated with single photon is as follows:

E=hcλ (3)

Substitute 2.5×10−7 m for λ , 6.626×10−34 J s for h and 3×108 m/s for c in equation (3).

E=(6.626×10−34 J s)(3×108 m/s)(2.5×10−7 m)=7.95×10−19 J/photon

Avogadro’s number is 6.022×1023 photons/mol. So energy for 1 mol photons is calculated as follows:

E=(6.022×1023 photons/mol)(7.95×10−19 J1 photon)(1 kJ1000 J)=478.7 kJ/mol≈479 kJ/mol

Thus, frequency, wavenumber and energy (J/photon and kJ/mol) are 1.2×1015 Hz, 4.0×103 cm−1, 7.95×10−19 J/photon and 479 kJ/mol respectively.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

Frequency, wavenumber and energy (J/photon and kJ/mol) of photon with 2.50 μm have to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Conversion factor to convert μm to m is as follows:

1 μm=10−6 m

Thus, 2.50 μm is converted to m as follows:

Wavelength=(2.50 μm )(10−6 m1 μm)=2.5×10−6 m

Substitute 2.5×10−6 m for λ , and 3×108 m/s for c in equation (1).

v=3×108 m/s2.5×10−6 m=1.2×1014 Hz

Substitute 1.2×1014 Hz for v , and 3×108 m/s for c in equation (2).

v¯=(1.2×1014 Hz3×108 m/s)(1 m102 cm)=4.0×103 cm−1

Substitute 2.5×10−6 m for λ , 6.626×10−34 J s for h and 3×108 m/s for c in equation (3).

E=(6.626×10−34 J s)(3×108 m/s)(2.5×10−6 m)=7.95×10−20 J/photon

Avogadro’s number is 6.022×1023 photons/mol. So energy for 1 mol photons is calculated as follows:

E=(6.022×1023 photons/mol)(7.95×10−20 J1 photon)(1 kJ1000 J)=47.87 kJ/mol≈47.9 kJ/mol

Thus, frequency, wavenumber and energy (J/photon and kJ/mol) are 1.2×1014 Hz, 4.0×103 cm−1, 7.95×10−20 J/photon and 47.9 kJ/mol respectively.

Answer 13

(b)

Interpretation Introduction

Interpretation:

Frequency, wavenumber and energy (J/photon and kJ/mol) of photon with 2.50 μm have to be calculated.

Concept Introduction:

Refer to part (a).

Answer 14

(b)

Expert Solution

Explanation of Solution

Conversion factor to convert μm to m is as follows:

1 μm=10−6 m

Thus, 2.50 μm is converted to m as follows:

Wavelength=(2.50 μm )(10−6 m1 μm)=2.5×10−6 m

Substitute 2.5×10−6 m for λ , and 3×108 m/s for c in equation (1).

v=3×108 m/s2.5×10−6 m=1.2×1014 Hz

Substitute 1.2×1014 Hz for v , and 3×108 m/s for c in equation (2).

v¯=(1.2×1014 Hz3×108 m/s)(1 m102 cm)=4.0×103 cm−1

Substitute 2.5×10−6 m for λ , 6.626×10−34 J s for h and 3×108 m/s for c in equation (3).

E=(6.626×10−34 J s)(3×108 m/s)(2.5×10−6 m)=7.95×10−20 J/photon

Avogadro’s number is 6.022×1023 photons/mol. So energy for 1 mol photons is calculated as follows:

E=(6.022×1023 photons/mol)(7.95×10−20 J1 photon)(1 kJ1000 J)=47.87 kJ/mol≈47.9 kJ/mol

Thus, frequency, wavenumber and energy (J/photon and kJ/mol) are 1.2×1014 Hz, 4.0×103 cm−1, 7.95×10−20 J/photon and 47.9 kJ/mol respectively.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

Ch. 18 - Prob. 18.1P Ch. 18 - Prob. 18.2P Ch. 18 - Prob. 18.3P Ch. 18 - Prob. 18.4P Ch. 18 - Prob. 18.5P Ch. 18 - Prob. 18.6P Ch. 18 - Prob. 18.7P Ch. 18 - Prob. 18.8P Ch. 18 - Prob. 18.9P Ch. 18 - Prob. 18.10P

Frequency, wavenumber and energy ( J/photon and kJ/mol ) of photon with 250 nm have to be calculated. Concept Introduction: Planck’s hypothesis states energy associated with photon is as follows: E = h c λ Here, λ denotes wavelength. h denotes Planck’s constant c denotes speed of light.

Explanation of Solution

Interpretation: Frequency, wavenumber and energy (J/photon and kJ/mol) of photon with 2.50 μm have to be calculated. Concept Introduction: Refer to part (a).

Explanation of Solution

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Chapter 18 Solutions

Interpretation:

Frequency, wavenumber and energy (J/photon and kJ/mol) of photon with 2.50 μm have to be calculated.

Concept Introduction:

Refer to part (a).