An electron in H-atom in its ground state absorbs ` 1.5` times as much energy as the minimum required for its escape ( i. e., 13 . 6 eV) from the atom . Calculate the wavelength of emitted electron.

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Determine the total energy absorbed by the electron
The minimum energy required for an electron to escape from the hydrogen atom is given as 13.6 eV. The electron absorbs 1.5 times this energy.

\[ \text{Total energy absorbed} = 1.5 \times 13.6 \, \text{eV} = 20.4 \, \text{eV} \]

### Step 2: Calculate the kinetic energy of the emitted electron
The kinetic energy (KE) of the emitted electron can be calculated by subtracting the energy required for escape from the total energy absorbed.

\[ \text{Kinetic Energy} = \text{Total energy absorbed} - \text{Escape energy} \]
\[ \text{Kinetic Energy} = 20.4 \, \text{eV} - 13.6 \, \text{eV} = 6.8 \, \text{eV} \]

### Step 3: Convert kinetic energy from eV to joules
To convert the kinetic energy from electron volts to joules, we use the conversion factor \(1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}\).

\[ \text{Kinetic Energy in Joules} = 6.8 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} = 1.090 \times 10^{-18} \, \text{J} \]

### Step 4: Use the kinetic energy to find the velocity of the electron
The kinetic energy can also be expressed in terms of mass and velocity:

\[ \text{KE} = \frac{1}{2} mv^2 \]

Rearranging for \(v\):

\[ v = \sqrt{\frac{2 \times \text{KE}}{m}} \]

Where the mass of the electron \(m = 9.1 \times 10^{-31} \, \text{kg}\).

Substituting the values:

\[ v = \sqrt{\frac{2 \times 1.090 \times 10^{-18} \, \text{J}}{9.1 \times 10^{-31} \, \text{kg}}} \]

Calculating this gives:

\[ v \approx 1.54 \times 10^8 \, \text{m/s} \]

### Step 5: Calculate the wavelength of the emitted electron
Using the de Broglie wavelength formula:

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

Where \(h = 6.626 \times 10^{-34} \, \text{J s}\).

Substituting the values:

\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{J s}}{(9.1 \times 10^{-31} \, \text{kg}) \times (1.54 \times 10^8 \, \text{m/s})} \]

Calculating this gives:

\[ \lambda \approx 4.72 \times 10^{-8} \, \text{m} = 4.72 \times 10^{-6} \, \text{cm} \]

### Final Answer
The wavelength of the emitted electron is approximately \(4.72 \times 10^{-8} \, \text{m}\) or \(4.72 \, \text{Å}\).

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