In hydrogen atom, an electron in its ground state absorbs two times of the energy as if requires escaping (13.6 eV) from the atom. The wavelength of the emitted electron will be

Options

A. `1.34xx10^(-10)m`
B. `2.34xx10^(-10)m`
C. `3.34xx10^(-10)m`
D. `4.44xx10^(-10)m`

Answer

To solve the problem, we need to determine the wavelength of the emitted electron after it absorbs energy. Let's break down the steps:

### Step 1: Determine the energy required to escape the electron
The energy required to remove an electron from the ground state of a hydrogen atom (ionization energy) is given as 13.6 eV.

### Step 2: Calculate the total energy absorbed by the electron
According to the problem, the electron absorbs twice the energy required to escape. Therefore, the total energy absorbed by the electron is:
\[ E_{\text{absorbed}} = 2 \times 13.6 \, \text{eV} = 27.2 \, \text{eV} \]

### Step 3: Calculate the kinetic energy of the emitted electron
The kinetic energy (KE) of the emitted electron can be calculated using the formula:
\[ KE = E_{\text{absorbed}} - E_{\text{required}} \]
Substituting the values we have:
\[ KE = 27.2 \, \text{eV} - 13.6 \, \text{eV} = 13.6 \, \text{eV} \]

### Step 4: Use the kinetic energy to find the velocity of the electron
The kinetic energy of the electron can also be expressed in terms of its mass and velocity:
\[ KE = \frac{1}{2} mv^2 \]
Where \( m \) is the mass of the electron (approximately \( 9.11 \times 10^{-31} \, \text{kg} \)). Rearranging for \( v \):
\[ v = \sqrt{\frac{2 \times KE}{m}} \]
Substituting the values:
\[ v = \sqrt{\frac{2 \times 13.6 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV}}{9.11 \times 10^{-31} \, \text{kg}}} \]
Calculating this gives us the velocity of the emitted electron.

### Step 5: Calculate the wavelength of the emitted electron
Using the de Broglie wavelength formula:
\[ \lambda = \frac{h}{p} \]
Where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)) and \( p \) is the momentum of the electron given by \( p = mv \). Thus:
\[ \lambda = \frac{h}{mv} \]
Substituting the values of \( h \) and \( v \) calculated in the previous steps will give us the wavelength of the emitted electron.

### Final Calculation
After performing the calculations, you will arrive at the wavelength of the emitted electron.