The de-Broglie wavlength of an electron emitted fromt the ground state of an H-atom after the absoprtion of a photon quals `(1)/(2)` of the de-Broglie wavelength when it was in orbit. The energy of the photon abserobed is

Options

A. `66 eV`
B. `13.6 eV`
C. `20.4 eV`
D. `136 eV`

Answer

To solve the problem, we need to find the energy of the photon absorbed by the electron in the ground state of a hydrogen atom, given that the de-Broglie wavelength of the emitted electron is half of its de-Broglie wavelength when it was in the atom.

### Step 1: Determine the energy of the electron in the ground state
The energy of an electron in the nth orbit of a hydrogen atom is given by the formula:

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

For the ground state (n = 1):

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

### Step 2: Calculate the energy required to free the electron
To free the electron from the atom, we need to provide enough energy to overcome its binding energy. This means we need to provide energy equal to the absolute value of the energy of the electron in the ground state:

\[ \text{Required energy} = 13.6 \, \text{eV} \]

### Step 3: Relate the de-Broglie wavelengths
Let’s denote the de-Broglie wavelength of the electron in the ground state as \(\lambda_1\) and the de-Broglie wavelength of the emitted electron as \(\lambda_2\). According to the problem, we have:

\[ \lambda_2 = \frac{1}{2} \lambda_1 \]

The de-Broglie wavelength is given by:

\[ \lambda = \frac{h}{p} \]

where \(h\) is Planck's constant and \(p\) is the momentum of the electron. The momentum of the electron in the ground state can be calculated using the energy relation:

\[ E = \frac{p^2}{2m} \implies p = \sqrt{2mE} \]

For the ground state:

\[ p_1 = \sqrt{2m \cdot 13.6 \, \text{eV}} \]

### Step 4: Calculate the de-Broglie wavelength in the ground state
Substituting \(p_1\) into the de-Broglie wavelength formula for the ground state:

\[ \lambda_1 = \frac{h}{p_1} = \frac{h}{\sqrt{2m \cdot 13.6 \, \text{eV}}} \]

### Step 5: Calculate the de-Broglie wavelength of the emitted electron
The emitted electron will have a different momentum, and since \(\lambda_2 = \frac{1}{2} \lambda_1\):

\[ \lambda_2 = \frac{h}{p_2} = \frac{1}{2} \lambda_1 \]

This implies:

\[ p_2 = 2p_1 \]

### Step 6: Determine the energy of the emitted electron
The kinetic energy of the emitted electron can be expressed as:

\[ \frac{p_2^2}{2m} = \frac{(2p_1)^2}{2m} = 2 \cdot \frac{p_1^2}{2m} \]

Since \(E_1 = \frac{p_1^2}{2m}\):

\[ \text{Kinetic Energy of emitted electron} = 2E_1 = 2 \cdot 13.6 \, \text{eV} = 27.2 \, \text{eV} \]

### Step 7: Calculate the total energy of the photon absorbed
The total energy of the photon absorbed must equal the energy required to free the electron plus the kinetic energy of the emitted electron:

\[ E_{\text{photon}} = 13.6 \, \text{eV} + 27.2 \, \text{eV} = 40.8 \, \text{eV} \]

### Final Answer
The energy of the photon absorbed is:

\[ \boxed{40.8 \, \text{eV}} \]