To solve this problem, we need to find the relative increase, [tex]\(\Delta W / W\)[/tex], in the number of accessible states of the system after it absorbs a photon with a wavelength of 500 nm. The steps to solve this include calculating the energy of the absorbed photon and then using this energy to find the relative increase in the number of accessible states. Below is a detailed, step-by-step solution.
### Step 1: Constants and Basic Information
1. Boltzmann constant ([tex]\(k_B\)[/tex]): [tex]\(1.380649 \times 10^{-23} \, \text{J/K}\)[/tex]
2. Temperature (T): [tex]\(300 \, \text{K}\)[/tex]
3. Planck's constant (h): [tex]\(6.62607015 \times 10^{-34} \, \text{J} \cdot \text{s}\)[/tex]
4. Speed of light (c): [tex]\(3 \times 10^8 \, \text{m/s}\)[/tex]
5. Wavelength of the photon ([tex]\(\lambda\)[/tex]): [tex]\(500 \times 10^{-9} \, \text{m}\)[/tex] (or 500 nm)
### Step 2: Energy of the Photon
We utilize the relation for the energy of a photon based on its wavelength:
[tex]\[
E_{\text{photon}} = \frac{h \cdot c}{\lambda}
\][/tex]
Plugging in the constants:
[tex]\[
E_{\text{photon}} = \frac{6.62607015 \times 10^{-34} \, \text{J} \cdot \text{s} \cdot 3 \times 10^8 \, \text{m/s}}{500 \times 10^{-9} \, \text{m}}
\][/tex]
### Step 3: Performing the Calculation
[tex]\[
E_{\text{photon}} = \frac{6.62607015 \times 10^{-34} \times 3 \times 10^8}{500 \times 10^{-9}}
\][/tex]
[tex]\[
E_{\text{photon}} = \frac{1.987821045 \times 10^{-25}}{500 \times 10^{-9}}
\][/tex]
[tex]\[
E_{\text{photon}} = \frac{1.987821045 \times 10^{-25}}{5 \times 10^{-7}}
\][/tex]
[tex]\[
E_{\text{photon}} = 3.97564209 \times 10^{-19} \, \text{J}
\][/tex]
### Step 4: Relative Increase in Number of Accessible States
The relative increase [tex]\(\Delta W / W\)[/tex] is given by the ratio of the photon's energy to the thermal energy ([tex]\(k_B \cdot T\)[/tex]) of the system:
[tex]\[
\Delta W / W \approx \frac{E_{\text{photon}}}{k_B \cdot T}
\][/tex]
Calculating [tex]\(k_B \cdot T\)[/tex]:
[tex]\[
k_B \cdot T = 1.380649 \times 10^{-23} \, \text{J/K} \times 300 \, \text{K} = 4.141947 \times 10^{-21} \, \text{J}
\][/tex]
Now, substituting in [tex]\(E_{\text{photon}}\)[/tex]:
[tex]\[
\Delta W / W \approx \frac{3.97564209 \times 10^{-19} \, \text{J}}{4.141947 \times 10^{-21} \, \text{J}}
\][/tex]
[tex]\[
\Delta W / W \approx 95.99
\][/tex]
### Conclusion
The relative increase, [tex]\(\Delta W / W\)[/tex], in the number of accessible states of the system after absorbing the photon with a wavelength of 500 nm at a temperature of 300 K is approximately 96.
