Westonci.ca offers fast, accurate answers to your questions. Join our community and get the insights you need now. Our platform provides a seamless experience for finding precise answers from a network of experienced professionals. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
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].
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.
