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### 6.1 Calculate the wavelength for an electron transition from [tex]\( n_{h} = 4 \)[/tex] to [tex]\( n_{1} = 1 \)[/tex] energy level in the hydrogen atom.
We use the Rydberg formula for this calculation:
[tex]\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_h^2} \right) \][/tex]
Here, the given parameters are:
- [tex]\( R = 1.1 \times 10^{-2} \, \text{nm}^{-1} \)[/tex]
- [tex]\( n_1 = 1 \)[/tex]
- [tex]\( n_h = 4 \)[/tex]
Plugging these values into the formula, we have:
[tex]\[ \frac{1}{\lambda} = 1.1 \times 10^{-2} \left( \frac{1}{1^2} - \frac{1}{4^2} \right) \][/tex]
Calculate the terms inside the parentheses:
[tex]\[ \frac{1}{1^2} = 1.0 \][/tex]
[tex]\[ \frac{1}{4^2} = \frac{1}{16} = 0.0625 \][/tex]
So,
[tex]\[ \frac{1}{\lambda} = 1.1 \times 10^{-2} \left( 1.0 - 0.0625 \right) = 1.1 \times 10^{-2} \times 0.9375 \][/tex]
Multiplying these values:
[tex]\[ \frac{1}{\lambda} = 0.0103125 \, \text{nm}^{-1} \][/tex]
Hence,
[tex]\[ \lambda = \frac{1}{0.0103125} \approx 96.97 \, \text{nm} \][/tex]
So, the wavelength ([tex]\(\lambda\)[/tex]) for the transition is approximately [tex]\( 96.97 \, \text{nm} \)[/tex].
### 6.2 Where (visible, ultraviolet, or infrared) does this line appear in the electromagnetic spectrum?
To determine the region of the electromagnetic spectrum where this line appears, we look at the calculated wavelength:
- Visible light: [tex]\(380\, \text{nm} - 750\, \text{nm}\)[/tex]
- Ultraviolet light: [tex]\( < 380\, \text{nm}\)[/tex]
- Infrared light: [tex]\( > 750\, \text{nm}\)[/tex]
Since the calculated wavelength is [tex]\( 96.97 \, \text{nm} \)[/tex], which is less than [tex]\( 380 \, \text{nm} \)[/tex], it falls in the ultraviolet region.
Thus, the line appears in the ultraviolet region.
### 6.3 Calculate the energy of the photon resulting from this electron transition.
The energy of the photon can be calculated using the equation:
[tex]\[ E = \frac{hc}{\lambda} \][/tex]
Where:
- [tex]\( h \)[/tex] is Planck's constant: [tex]\( 6.626 \times 10^{-34} \, \text{m}^2 \, \text{kg} / \text{s} \)[/tex]
- [tex]\( c \)[/tex] is the speed of light: [tex]\( 3 \times 10^8 \, \text{m} / \text{s} \)[/tex]
- [tex]\( \lambda \)[/tex] is the wavelength: [tex]\( 96.97 \, \text{nm} = 96.97 \times 10^{-9} \, \text{m} \)[/tex]
Plug these values into the equation:
[tex]\[ E = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{96.97 \times 10^{-9}} \][/tex]
Calculating the numerator:
[tex]\[ 6.626 \times 10^{-34} \times 3 \times 10^8 = 1.9878 \times 10^{-25} \][/tex]
Now, divide by the denominator:
[tex]\[ E = \frac{1.9878 \times 10^{-25}}{96.97 \times 10^{-9}} \approx 2.0499 \times 10^{-18} \, \text{J} \][/tex]
Therefore, the energy of the photon resulting from this electron transition is approximately [tex]\( 2.0499 \times 10^{-18} \, \text{J} \)[/tex].
In summary:
1. The wavelength of the transition is approximately [tex]\( 96.97 \, \text{nm} \)[/tex].
2. This line appears in the ultraviolet region.
3. The energy of the photon resulting from this transition is approximately [tex]\( 2.0499 \times 10^{-18} \, \text{J} \)[/tex].
### 6.1 Calculate the wavelength for an electron transition from [tex]\( n_{h} = 4 \)[/tex] to [tex]\( n_{1} = 1 \)[/tex] energy level in the hydrogen atom.
We use the Rydberg formula for this calculation:
[tex]\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_h^2} \right) \][/tex]
Here, the given parameters are:
- [tex]\( R = 1.1 \times 10^{-2} \, \text{nm}^{-1} \)[/tex]
- [tex]\( n_1 = 1 \)[/tex]
- [tex]\( n_h = 4 \)[/tex]
Plugging these values into the formula, we have:
[tex]\[ \frac{1}{\lambda} = 1.1 \times 10^{-2} \left( \frac{1}{1^2} - \frac{1}{4^2} \right) \][/tex]
Calculate the terms inside the parentheses:
[tex]\[ \frac{1}{1^2} = 1.0 \][/tex]
[tex]\[ \frac{1}{4^2} = \frac{1}{16} = 0.0625 \][/tex]
So,
[tex]\[ \frac{1}{\lambda} = 1.1 \times 10^{-2} \left( 1.0 - 0.0625 \right) = 1.1 \times 10^{-2} \times 0.9375 \][/tex]
Multiplying these values:
[tex]\[ \frac{1}{\lambda} = 0.0103125 \, \text{nm}^{-1} \][/tex]
Hence,
[tex]\[ \lambda = \frac{1}{0.0103125} \approx 96.97 \, \text{nm} \][/tex]
So, the wavelength ([tex]\(\lambda\)[/tex]) for the transition is approximately [tex]\( 96.97 \, \text{nm} \)[/tex].
### 6.2 Where (visible, ultraviolet, or infrared) does this line appear in the electromagnetic spectrum?
To determine the region of the electromagnetic spectrum where this line appears, we look at the calculated wavelength:
- Visible light: [tex]\(380\, \text{nm} - 750\, \text{nm}\)[/tex]
- Ultraviolet light: [tex]\( < 380\, \text{nm}\)[/tex]
- Infrared light: [tex]\( > 750\, \text{nm}\)[/tex]
Since the calculated wavelength is [tex]\( 96.97 \, \text{nm} \)[/tex], which is less than [tex]\( 380 \, \text{nm} \)[/tex], it falls in the ultraviolet region.
Thus, the line appears in the ultraviolet region.
### 6.3 Calculate the energy of the photon resulting from this electron transition.
The energy of the photon can be calculated using the equation:
[tex]\[ E = \frac{hc}{\lambda} \][/tex]
Where:
- [tex]\( h \)[/tex] is Planck's constant: [tex]\( 6.626 \times 10^{-34} \, \text{m}^2 \, \text{kg} / \text{s} \)[/tex]
- [tex]\( c \)[/tex] is the speed of light: [tex]\( 3 \times 10^8 \, \text{m} / \text{s} \)[/tex]
- [tex]\( \lambda \)[/tex] is the wavelength: [tex]\( 96.97 \, \text{nm} = 96.97 \times 10^{-9} \, \text{m} \)[/tex]
Plug these values into the equation:
[tex]\[ E = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{96.97 \times 10^{-9}} \][/tex]
Calculating the numerator:
[tex]\[ 6.626 \times 10^{-34} \times 3 \times 10^8 = 1.9878 \times 10^{-25} \][/tex]
Now, divide by the denominator:
[tex]\[ E = \frac{1.9878 \times 10^{-25}}{96.97 \times 10^{-9}} \approx 2.0499 \times 10^{-18} \, \text{J} \][/tex]
Therefore, the energy of the photon resulting from this electron transition is approximately [tex]\( 2.0499 \times 10^{-18} \, \text{J} \)[/tex].
In summary:
1. The wavelength of the transition is approximately [tex]\( 96.97 \, \text{nm} \)[/tex].
2. This line appears in the ultraviolet region.
3. The energy of the photon resulting from this transition is approximately [tex]\( 2.0499 \times 10^{-18} \, \text{J} \)[/tex].
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