A photon with a wavelength of 93.8 nm strikes a hydrogen atom, and light is emitted by the atom. How many emission lines would be observed?
hc/wavelength = 2.180E-18(1/1 - 1/x^2)
Substitute for h, c, and wavelength (in meters for wavelength) and solve for x which will be the orbit to which the photon raised the electron in the ground hydrogen atom. Then count the lines it can emit from its excited state. Post your work if you get stuck.
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To determine the number of emission lines observed when a photon of a specific wavelength strikes a hydrogen atom, we need to consider the energy levels of the hydrogen atom and the possible transitions between them.
The energy levels of a hydrogen atom are given by the Rydberg formula:
\[ E = -\frac{{13.6 \, \text{{eV}}}}{{n^2}} \]
where E is the energy in electron volts (eV) and n is the principal quantum number of the energy level.
The energy change between two energy levels can be calculated using the formula:
\[ \Delta E = E_{\text{{final}}} - E_{\text{{initial}}} \]
For hydrogen, the initial energy level (n_1) is typically the ground state with principal quantum number n = 1.
The wavelength of the emitted light can be calculated using the relationship:
\[ \text{{wavelength}} = \frac{{\text{{speed of light}}}}{{\text{{frequency}}}} \]
The frequency of the emitted light can be calculated using the relationship:
\[ \text{{frequency}} = \frac{{\Delta E}}{{h}} \]
where h is the Planck's constant.
To calculate the number of emission lines, we need to find the possible transitions that result in the given wavelength of the emitting photon.
1. Convert the given wavelength from nm to meters:
93.8 nm = 93.8 × 10^(-9) m
2. Calculate the frequency of the emitted light:
\[ \text{{frequency}} = \frac{{\text{{speed of light}}}}{{\text{{wavelength}}}} \]
3. Calculate the energy change (ΔE) using the relationship:
\[ \Delta E = h \times \text{{frequency}} \]
4. Calculate the energy levels (E_final) using the Rydberg formula:
\[ E_{\text{{final}}} = -\frac{{13.6 \, \text{{eV}}}}{{n^2}} \]
5. Determine the principal quantum number (n) for each transition by rearranging the Rydberg formula:
\[ n = \sqrt{\frac{{-13.6 \, \text{{eV}}}}{{E_{\text{{final}}}}}} \]
6. Check if the calculated principal quantum number (n) is an integer for each transition. If it is, then that transition is possible.
7. Count the number of possible transitions, which will give us the number of emission lines observed.
Please note that the above approach assumes the hydrogen atom is in its ground state and does not consider any external influences that may affect the energy levels and transitions. It also assumes the single-photon interaction with the atom, neglecting any multi-photon processes.
To determine the number of emission lines observed when a photon strikes a hydrogen atom, we need to understand the energy levels of hydrogen and how emission lines are produced.
Hydrogen atom emission lines arise from the transitions of electrons between different energy levels within the atom. When an electron moves from a higher energy level to a lower one, it releases energy in the form of photons, which correspond to specific wavelengths of light.
Each energy level in hydrogen is associated with a specific whole number value called the principal quantum number, represented by the symbol "n." The smallest value of n is 1, corresponding to the ground state, and higher energy levels have larger values of n.
The energy of a photon can be calculated using the equation:
E = hc/λ
Where:
E is the energy of the photon
h is Planck's constant (6.626 x 10^-34 J·s)
c is the speed of light in a vacuum (approximately 3.00 x 10^8 m/s)
λ is the wavelength of the photon
We can use this equation to calculate the energy of the given photon with a wavelength of 93.8 nm.
First, we need to convert the wavelength from nanometers to meters:
93.8 nm = 93.8 x 10^-9 m
Now, we can calculate the energy of the photon:
E = (6.626 x 10^-34 J·s) x (3.00 x 10^8 m/s) / (93.8 x 10^-9 m)
By performing this calculation, we find that the energy of the photon is approximately 2.11 x 10^-18 J.
To determine the energy levels involved in the emission lines, we need to consider the energy differences between the involved levels. In hydrogen, the energy difference between energy levels can be determined using the Rydberg formula:
ΔE = -13.6 eV (1/nf^2 - 1/ni^2)
Where:
ΔE is the energy difference between the levels
-13.6 eV is the ionization energy of hydrogen (the energy required to free the electron completely)
ni and nf are the initial and final energy levels, respectively
For the emission lines to be observed, the energy difference between the levels involved must match the energy of the photon striking the atom (approximately 2.11 x 10^-18 J). Now, we can check which energy differences match this value.
Considering the energy differences between the ground state (n=1) and different higher energy levels (n=2, 3, 4...), we can calculate the corresponding wavelengths of light using the equation:
λ = hc / ΔE
By substituting the known values for h, c, and ΔE, we can calculate the wavelengths of light emitted for each energy level difference.
To find the number of emission lines observed, we count the number of unique wavelengths obtained. Each unique wavelength corresponds to a distinct emission line.
Note that only transitions from higher energy levels to the lower energy level (n=1) will result in the observed emission lines.
By performing these calculations, we would find the number of emission lines that would be observed when a photon with a wavelength of 93.8 nm strikes a hydrogen atom.