An electron joins a helium nucleus to form `He^(+)`. Find the wavelength of the photon emitted in this process (in `nm`) if the electron is assumed to have no kinetic energy when it combines with nucleus. Assume Bhor's nodel to be applicable.

Explanation

To find the wavelength of the photon emitted when an electron joins a helium nucleus to form \( \text{He}^+ \), we can follow these steps:

### Step 1: Determine the Energy of the Electron in \( \text{He}^+ \)
According to Bohr's model, the energy of an electron in a hydrogen-like atom is given by the formula:

\[ E_n = -\frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]

For \( \text{He}^+ \), the atomic number \( Z = 2 \) and we will consider the ground state where \( n = 1 \).

Substituting the values:

\[ E_1 = -\frac{13.6 \, \text{eV} \cdot 2^2}{1^2} = -\frac{13.6 \, \text{eV} \cdot 4}{1} = -54.4 \, \text{eV} \]

### Step 2: Calculate the Energy of the Photon Emitted
Since the electron is assumed to have no kinetic energy when it combines with the nucleus, the energy emitted as a photon when the electron transitions from infinity (where the energy is 0 eV) to the ground state of \( \text{He}^+ \) is equal to the absolute value of the energy of the electron in the ground state:

\[ E_{\text{photon}} = 54.4 \, \text{eV} \]

### Step 3: Use the Energy-Wavelength Relationship
The energy of a photon is related to its wavelength \( \lambda \) by the equation:

\[ E = \frac{hc}{\lambda} \]

Where:
- \( h = 6.626 \times 10^{-34} \, \text{J s} \) (Planck's constant)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)

### Step 4: Convert Energy from eV to Joules
To use the above formula, we need to convert the energy from electron volts to joules. The conversion factor is:

\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]

Thus,

\[ E_{\text{photon}} = 54.4 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 8.704 \times 10^{-18} \, \text{J} \]

### Step 5: Rearranging the Energy-Wavelength Equation
Rearranging the equation for wavelength gives:

\[ \lambda = \frac{hc}{E} \]

Substituting the values:

\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{8.704 \times 10^{-18} \, \text{J}} \]

### Step 6: Calculate the Wavelength
Calculating the above expression:

\[ \lambda = \frac{1.9878 \times 10^{-25}}{8.704 \times 10^{-18}} \approx 2.28 \times 10^{-8} \, \text{m} \]

Converting to nanometers (1 nm = \( 10^{-9} \, \text{m} \)):

\[ \lambda \approx 22.8 \, \text{nm} \]

### Final Answer
The wavelength of the photon emitted in this process is approximately \( 22.8 \, \text{nm} \).

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