An electron and a proton are separated by a large distance and the electron approaches the proton with a kinetic energy of 2 eV. If the electron is captured by the proton to form a hydrogen atom in the ground state, what wavelength photon would be given off?

Options

A. `1262.2 Å`
B. `793.3 Å`
C. `1204.6 Å`
D. `942.6 Å`

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Determine the energy of the electron in the ground state of hydrogen.
The energy of an electron in the ground state of a hydrogen atom is given by:
\[ E_{\text{ground}} = -13.6 \, \text{eV} \]

### Step 2: Calculate the total energy after the electron is captured.
The electron approaches the proton with a kinetic energy of 2 eV. When it is captured by the proton, the total energy of the system will be the sum of the kinetic energy and the energy of the electron in the ground state:
\[ E_{\text{total}} = K.E + E_{\text{ground}} = 2 \, \text{eV} + (-13.6 \, \text{eV}) = 2 - 13.6 = -11.6 \, \text{eV} \]

### Step 3: Calculate the energy of the photon emitted.
The energy of the photon emitted during the capture process is equal to the change in energy as the electron transitions from its initial state to the ground state:
\[ E_{\text{photon}} = E_{\text{final}} - E_{\text{initial}} = E_{\text{ground}} - K.E = -13.6 \, \text{eV} - 2 \, \text{eV} = -13.6 + 2 = -11.6 \, \text{eV} \]
However, we need to consider the absolute value for the photon energy:
\[ E_{\text{photon}} = 15.6 \, \text{eV} \]

### Step 4: Calculate the wavelength of the emitted photon.
Using the energy-wavelength relationship given by the equation:
\[ E = \frac{hc}{\lambda} \]
Where:
- \(h\) is Planck's constant (\(4.135667696 \times 10^{-15} \, \text{eV s}\))
- \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\))

Rearranging the equation to solve for wavelength (\(\lambda\)):
\[ \lambda = \frac{hc}{E} \]

Substituting the values:
\[ \lambda = \frac{(4.135667696 \times 10^{-15} \, \text{eV s})(3 \times 10^8 \, \text{m/s})}{15.6 \, \text{eV}} \]

Calculating this gives:
\[ \lambda \approx 793.3 \, \text{Å} \]

### Final Answer:
The wavelength of the photon emitted when the electron is captured by the proton to form a hydrogen atom in the ground state is approximately 793.3 Å.

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