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. `2.4 eV`
B. `2.7 eV`
C. `2.9 eV`
D. `5.4 eV`

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 the electron in the ground state of a hydrogen atom is given by the formula:
\[ E = -13.6 \, \text{eV} \]
This value indicates that the electron is bound to the proton.

### 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 (E_total) can be calculated as:
\[ E_{\text{total}} = \text{Kinetic Energy} + \text{Energy of Ground State} \]
Substituting the values:
\[ E_{\text{total}} = 2 \, \text{eV} + (-13.6 \, \text{eV}) = 2 - 13.6 = -11.6 \, \text{eV} \]

### Step 3: Calculate the energy of the photon emitted
When the electron is captured, it transitions from a state with higher energy (kinetic energy) to a lower energy state (ground state). The energy of the photon emitted (E_photon) during this transition can be calculated as:
\[ E_{\text{photon}} = \text{Initial Energy} - \text{Final Energy} \]
Here, the initial energy is the kinetic energy of the electron (2 eV) and the final energy is the energy of the electron in the ground state (-13.6 eV):
\[ E_{\text{photon}} = 2 \, \text{eV} - (-13.6 \, \text{eV}) = 2 + 13.6 = 15.6 \, \text{eV} \]

### Step 4: Calculate the wavelength of the emitted photon
The wavelength (λ) of the emitted photon can be calculated using the energy of the photon and the formula:
\[ E = \frac{hc}{\lambda} \]
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 formula gives:
\[ \lambda = \frac{hc}{E} \]
Substituting the values:
\[ \lambda = \frac{(4.1357 \times 10^{-15} \, \text{eV s})(3 \times 10^8 \, \text{m/s})}{15.6 \, \text{eV}} \]
Calculating this gives:
\[ \lambda = \frac{1.24071 \times 10^{-6} \, \text{m}}{15.6} \approx 7.93 \times 10^{-7} \, \text{m} = 793.3 \, \text{nm} \]

### 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:
\[ \lambda \approx 793.3 \, \text{nm} \]