An electron and a proton are seperated by a large distance and the electorn approaches the proton with a kinectic 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?

Explanation

To solve the problem, we need to determine the wavelength of the photon emitted when an electron is captured by a proton to form a hydrogen atom in the ground state. Here’s the step-by-step solution:

### Step 1: Determine the energy of the hydrogen atom in the ground state.
The energy of a hydrogen atom in the ground state (n=1) is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
For n=1:
\[ E_1 = -13.6 \, \text{eV} \]

### Step 2: Calculate the total energy before the electron is captured.
The total energy before the capture is the kinetic energy of the electron, which is given as 2 eV. Since the proton is initially at rest, we can consider its energy to be 0 eV. Thus, the total initial energy (E_initial) is:
\[ E_{\text{initial}} = 2 \, \text{eV} + 0 \, \text{eV} = 2 \, \text{eV} \]

### Step 3: Calculate the energy released during the capture.
When the electron is captured by the proton, it transitions from a state with energy of 2 eV to the ground state of the hydrogen atom, which has an energy of -13.6 eV. The energy released (E_released) during this process is:
\[ E_{\text{released}} = E_{\text{final}} - E_{\text{initial}} \]
\[ E_{\text{released}} = (-13.6 \, \text{eV}) - (2 \, \text{eV}) \]
\[ E_{\text{released}} = -15.6 \, \text{eV} \]

### Step 4: Calculate the energy of the emitted photon.
The energy of the emitted photon is equal to the absolute value of the energy released:
\[ E_{\text{photon}} = 15.6 \, \text{eV} \]

### Step 5: Use the energy of the photon to find its wavelength.
The energy of a photon is related to its wavelength (λ) by the equation:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h \) is Planck's constant \( (4.1357 \times 10^{-15} \, \text{eV s}) \)
- \( c \) is the speed of light \( (3 \times 10^8 \, \text{m/s}) \)

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

### Step 6: Substitute the values to find the wavelength.
Using the value of \( hc \) in eV·nm (1.24 x 10^-6 eV·m):
\[ \lambda = \frac{1.24 \times 10^{-6} \, \text{eV m}}{15.6 \, \text{eV}} \]
\[ \lambda \approx 7.95 \times 10^{-8} \, \text{m} \]
Converting to angstroms (1 m = 10^10 angstroms):
\[ \lambda \approx 793 \, \text{angstroms} \]

### Final Answer:
The wavelength of the photon emitted is approximately **793 angstroms**.

---