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To find the wavelength of light required to excite an electron in a hydrogen atom from level \( n = 1 \) to \( n = 2 \), we will follow these steps:
1. Calculate the energy of the electron at the initial and final energy levels.
The energy of an electron in a hydrogen atom can be calculated using the formula:
[tex]\[ E_n = -2.178 \times 10^{-18} \left(\frac{z^2}{n^2}\right) \text{J} \][/tex]
Here, \( z \) is the atomic number (for hydrogen, \( z = 1 \)), and \( n \) is the principal quantum number.
For the initial energy level \( n_1 = 1 \):
[tex]\[ E_{initial} = -2.178 \times 10^{-18} \left(\frac{1^2}{1^2}\right) = -2.178 \times 10^{-18} \text{J} \][/tex]
For the final energy level \( n_2 = 2 \):
[tex]\[ E_{final} = -2.178 \times 10^{-18} \left(\frac{1^2}{2^2}\right) = -2.178 \times 10^{-18} \left(\frac{1}{4}\right) = -8.712 \times 10^{-19} \text{J} \][/tex]
Simplification of the given constants confirms this value:
[tex]\[ -8.712 \times 10^{-18} \text{J} \][/tex]
2. Calculate the energy difference (\( \Delta E \)) between the initial and final energy levels.
[tex]\[ \Delta E = E_{final} - E_{initial} \][/tex]
Substituting the values:
[tex]\[ \Delta E = -8.712 \times 10^{-18} \text{J} - (-2.178 \times 10^{-18} \text{J}) \][/tex]
[tex]\[ \Delta E = -8.712 \times 10^{-18} \text{J} + 2.178 \times 10^{-18} \text{J} \][/tex]
[tex]\[ \Delta E = -6.534 \times 10^{-18} \text{J} \][/tex]
3. Calculate the wavelength of the light (\( \lambda \)) using the energy difference.
The relationship between the energy difference and the wavelength of light is given by the equation:
[tex]\[ E = \frac{h \cdot c}{\lambda} \][/tex]
Rearranging this to solve for \( \lambda \):
[tex]\[ \lambda = \frac{h \cdot c}{| \Delta E |} \][/tex]
Substituting the values \( h = 6.62 \times 10^{-34} \text{Js} \), \( c = 3.0 \times 10^8 \text{m/s} \), and \( \Delta E = -6.534 \times 10^{-18} \text{J} \):
[tex]\[ \lambda = \frac{6.62 \times 10^{-34} \text{Js} \cdot 3.0 \times 10^8 \text{m/s}}{6.534 \times 10^{-18} \text{J}} \][/tex]
This results in:
[tex]\[ \lambda \approx 3.039 \times 10^{-8} \text{m} \][/tex]
Further simplifying and converting, we get:
[tex]\[ \lambda = 3.039 \times 10^{-8} \text{m} \][/tex]
Hence, the wavelength of light required to excite an electron in a hydrogen atom from level [tex]\( n = 1 \)[/tex] to [tex]\( n = 2 \)[/tex] is approximately [tex]\( 3.039 \times 10^{-8} \text{m} \)[/tex].
1. Calculate the energy of the electron at the initial and final energy levels.
The energy of an electron in a hydrogen atom can be calculated using the formula:
[tex]\[ E_n = -2.178 \times 10^{-18} \left(\frac{z^2}{n^2}\right) \text{J} \][/tex]
Here, \( z \) is the atomic number (for hydrogen, \( z = 1 \)), and \( n \) is the principal quantum number.
For the initial energy level \( n_1 = 1 \):
[tex]\[ E_{initial} = -2.178 \times 10^{-18} \left(\frac{1^2}{1^2}\right) = -2.178 \times 10^{-18} \text{J} \][/tex]
For the final energy level \( n_2 = 2 \):
[tex]\[ E_{final} = -2.178 \times 10^{-18} \left(\frac{1^2}{2^2}\right) = -2.178 \times 10^{-18} \left(\frac{1}{4}\right) = -8.712 \times 10^{-19} \text{J} \][/tex]
Simplification of the given constants confirms this value:
[tex]\[ -8.712 \times 10^{-18} \text{J} \][/tex]
2. Calculate the energy difference (\( \Delta E \)) between the initial and final energy levels.
[tex]\[ \Delta E = E_{final} - E_{initial} \][/tex]
Substituting the values:
[tex]\[ \Delta E = -8.712 \times 10^{-18} \text{J} - (-2.178 \times 10^{-18} \text{J}) \][/tex]
[tex]\[ \Delta E = -8.712 \times 10^{-18} \text{J} + 2.178 \times 10^{-18} \text{J} \][/tex]
[tex]\[ \Delta E = -6.534 \times 10^{-18} \text{J} \][/tex]
3. Calculate the wavelength of the light (\( \lambda \)) using the energy difference.
The relationship between the energy difference and the wavelength of light is given by the equation:
[tex]\[ E = \frac{h \cdot c}{\lambda} \][/tex]
Rearranging this to solve for \( \lambda \):
[tex]\[ \lambda = \frac{h \cdot c}{| \Delta E |} \][/tex]
Substituting the values \( h = 6.62 \times 10^{-34} \text{Js} \), \( c = 3.0 \times 10^8 \text{m/s} \), and \( \Delta E = -6.534 \times 10^{-18} \text{J} \):
[tex]\[ \lambda = \frac{6.62 \times 10^{-34} \text{Js} \cdot 3.0 \times 10^8 \text{m/s}}{6.534 \times 10^{-18} \text{J}} \][/tex]
This results in:
[tex]\[ \lambda \approx 3.039 \times 10^{-8} \text{m} \][/tex]
Further simplifying and converting, we get:
[tex]\[ \lambda = 3.039 \times 10^{-8} \text{m} \][/tex]
Hence, the wavelength of light required to excite an electron in a hydrogen atom from level [tex]\( n = 1 \)[/tex] to [tex]\( n = 2 \)[/tex] is approximately [tex]\( 3.039 \times 10^{-8} \text{m} \)[/tex].
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