A photon of wavelength `12.4 nm `.is used to eject an electron from the ground state of `He^+ .` The de-Broglie waveIength of the ejected electron is `lambda_(b) A^(circ)` .` Find `lambda_(delta)^(2) .` (Round off answer to two places of decimal.)

Answer

To solve the problem step by step, we need to find the de Broglie wavelength of the ejected electron when a photon of wavelength 12.4 nm is used to eject it from the ground state of \( He^+ \).

### Step 1: Calculate the energy of the photon
The energy \( E \) of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h = 4.1357 \times 10^{-15} \) eV·s (Planck's constant)
- \( c = 3 \times 10^{10} \) cm/s (speed of light)
- \( \lambda = 12.4 \) nm = \( 12.4 \times 10^{-9} \) m = \( 12.4 \) nm

Substituting these values:
\[ E = \frac{1240 \text{ eV·nm}}{12.4 \text{ nm}} = 100 \text{ eV} \]

### Step 2: Calculate the energy of the electron in the ground state of \( He^+ \)
The energy of an electron in the ground state of a hydrogen-like atom is given by:
\[ E_n = -\frac{Z^2 \cdot 13.6 \text{ eV}}{n^2} \]
For \( He^+ \) (where \( Z = 2 \) and \( n = 1 \)):
\[ E_1 = -\frac{2^2 \cdot 13.6 \text{ eV}}{1^2} = -54.4 \text{ eV} \]

### Step 3: Calculate the kinetic energy of the ejected electron
The kinetic energy \( KE \) of the ejected electron can be found using:
\[ KE = E_{\text{photon}} - |E_{n}| \]
Substituting the values:
\[ KE = 100 \text{ eV} - 54.4 \text{ eV} = 45.6 \text{ eV} \]

### Step 4: Calculate the de Broglie wavelength of the ejected electron
The de Broglie wavelength \( \lambda_b \) is given by:
\[ \lambda_b = \frac{h}{p} \]
where \( p \) is the momentum of the electron. The momentum can be expressed in terms of kinetic energy:
\[ p = \sqrt{2m \cdot KE} \]
Using \( m \) (mass of the electron) \( \approx 9.11 \times 10^{-31} \) kg, we first convert the kinetic energy from eV to Joules:
\[ KE = 45.6 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 7.296 \times 10^{-18} \text{ J} \]
Now, calculate the momentum:
\[ p = \sqrt{2 \cdot (9.11 \times 10^{-31} \text{ kg}) \cdot (7.296 \times 10^{-18} \text{ J})} \approx \sqrt{1.327 \times 10^{-48}} \approx 1.152 \times 10^{-24} \text{ kg·m/s} \]

Now, substitute \( p \) back into the de Broglie wavelength formula:
\[ \lambda_b = \frac{h}{p} = \frac{4.1357 \times 10^{-15} \text{ eV·s}}{1.152 \times 10^{-24} \text{ kg·m/s}} \approx 3.59 \times 10^{-10} \text{ m} = 3.59 \text{ Å} \]

### Step 5: Calculate \( \lambda_{\delta}^2 \)
To find \( \lambda_{\delta}^2 \):
\[ \lambda_{\delta}^2 = (3.59 \text{ Å})^2 \approx 12.88 \text{ Å}^2 \]
Rounding off to two decimal places, we have:
\[ \lambda_{\delta}^2 \approx 12.88 \]

### Final Answer:
\[ \lambda_{\delta}^2 \approx 12.88 \]