Concept explainers
(a)
Interpretation:
The energy density in the range 650 nm to 655 nm inside a cavity at 25° C has to be calculated.
Concept introduction:
Energy density is defined as the total energy inside the container divided by its volume. The energy density at any temperature, T due to the presence of the
(a)
Answer to Problem 7A.1P
The energy density in the range 650 nm to 655 nm inside a cavity at 25° C is 1.42× 10−33 J m−3_.
Explanation of Solution
The formula for Planck’s distribution is,
ρ(λ,T)=8πhcλ5(ehc/λkT−1) (1)
Where,
- λ is the wavelength.
- h is the Planck’s constant (6.626×10−34 kg m2/s).
- c is the
speed of light (3×108 m/s). - T is the temperature.
- ρ is the energy spectral density.
- k is the Boltzmann’s distribution constant (1.38×10−23 kg m2 s−2).
The range of wavelength (Δλ) is,
Δλ=655 nm−650 nm=5 nm
The conversion of nm to m is done as,
1 nm= 10−9 m
Therefore, the conversion of 5 nm to m is done as,
5 nm=5× 10−9 m
The given range of wavelength is small. Hence, the approximation can be used.
ΔE=ρΔλ (2)
Where,
- ΔE is the energy density.
- ρ is the energy spectral density.
- Δλ is the range of wavelength.
The value of λ is calculated as,
λ=655+6502=652.5 nm
The conversion of nm to m is done as,
1 nm= 10−9 m
Therefore, the conversion of 652.5 nm to m is done as,
652.5 nm=652.5 × 10−9 m
The value of λ is 652.5 × 10−9 m.
Substitute the value of h, c, k and λ in equation (1).
ρ(λ,T)=8×3.14×(6.626×10−34 kg m2/s)×(3×108 m/s)(652.5 × 10−9 m)5(e6.626×10−34×3×108/652.5 × 10−9 ×1.38×10−23×T−1)=4.99×10−241.18× 10−31(e2.2075×104/T−1)=4.2× 107(e2.2075×104/T−1)
The given temperature is 25° C.
The conversion of given temperature (°C) into K is shown below.
K=°C+273=25°C+273=298 K
The value of Δλ is 5× 10−9 m.
The value of ρ is 4.2× 107(e2.2075×104/T−1).
The value of T is 298 K.
Substitute the value of Δλ, T and ρ in equation (2).
ΔE=4.2× 107(e2.2075×104/298−1)×5× 10−9 m=4.2× 107(1.48× 1032−1)×5× 10−9 =0.211.48× 1032=1.42× 10−33 J m−3_
Hence, the energy density in the range 650 nm to 655 nm inside a cavity at 25° C is 1.42× 10−33 J m−3_.
(b)
Interpretation:
The energy density in the range 650 nm to 655 nm inside a cavity at 3000° C has to be calculated.
Concept introduction:
Energy density is defined as the total energy inside the container divided by its volume. The energy density at any temperature, T due to the presence of the electromagnetic radiations of the wavelength ranging between λ and λ+dλ is known as energy spectral density.
(b)
Answer to Problem 7A.1P
The energy density in the range 650 nm to 655 nm inside a cavity at 3000° C is 2.4× 10−4 J m−3_.
Explanation of Solution
The formula for Planck’s distribution is,
ρ(λ,T)=8πhcλ5(ehc/λkT−1) (1)
Where,
- λ is the wavelength.
- h is the Planck’s constant (6.626×10−34 kg m2/s).
- c is the speed of light (3×108 m/s).
- T is the temperature.
- ρ is the energy spectral density.
- k is the Boltzmann’s distribution constant (1.38×10−23 kg m2 s−2).
The range of wavelength (Δλ) is,
Δλ=655 nm−650 nm=5 nm
The conversion of nm to m is done as,
1 nm= 10−9 m
Therefore, the conversion of 5 nm to m is done as,
5 nm=5× 10−9 m
The given range of wavelength is small. Hence, the approximation can be used.
ΔE=ρΔλ (2)
Where,
- ΔE is the energy density.
- ρ is the energy spectral density.
- Δλ is the range of wavelength.
The value of λ is calculated as,
λ=655+6502=652.5 nm
The conversion of nm to m is done as,
1 nm= 10−9 m
Therefore, the conversion of 652.5 nm to m is done as,
652.5 nm=652.5 × 10−9 m
The value of λ is 652.5 × 10−9 m.
Substitute the value of h, c, k and λ in equation (1).
ρ(λ,T)=8×3.14×(6.626×10−34 kg m2/s)×(3×108 m/s)(652.5 × 10−9 m)5(e6.626×10−34×3×108/652.5 × 10−9 ×1.38×10−23×T−1)=4.99×10−241.18× 10−31(e2.2075×104/T−1)=4.2× 107(e2.2075×104/T−1)
The given temperature is 3000° C.
The conversion of given temperature (°C) into K is shown below.
K=°C+273=3000° C+273=3273 K
The value of Δλ is 5× 10−9 m.
The value of ρ is 4.2× 107(e2.2075×104/T−1).
The value of T is 3273 K.
Substitute the value of Δλ, T and ρ in equation (2).
ΔE=4.2× 107(e2.2075×104/3273−1)×5× 10−9 m=4.2× 107(8.48× 102)×5× 10−9 =0.21(8.48× 102)=2.4× 10−4 J m−3_
Hence, the energy density in the range 650 nm to 655 nm inside a cavity at 3000° C is 2.4× 10−4 J m−3_.
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Chapter 7 Solutions
Student Solutions Manual to accompany Atkins' Physical Chemistry 11th edition
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