The de - Broglie wavelength of an electron having `80 ev` of energy is nearly
`( 1eV = 1.6 xx 10^(-19) J`, Mass of electron ` = 9 xx 10^(-31) kg` Plank's constant ` = 6.6 xx 10^(-34) J - sec`)

Options

A. `140 Å`
B. `0.14 Å`
C. `14 Å`
D. `1.4 Å`

Explanation

To find the de Broglie wavelength of an electron with an energy of 80 eV, we can follow these steps:

### Step 1: Convert Energy from eV to Joules
The energy of the electron is given as 80 eV. We need to convert this energy into Joules using the conversion factor \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}\).

\[ E = 80 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 1.28 \times 10^{-17} \text{ J} \]

### Step 2: Calculate the Momentum of the Electron
The momentum \(P\) of the electron can be calculated using the formula:

\[ P = \sqrt{2mE} \]

Where:
- \(m\) is the mass of the electron \(= 9 \times 10^{-31} \text{ kg}\)
- \(E\) is the energy in Joules \(= 1.28 \times 10^{-17} \text{ J}\)

Substituting the values:

\[ P = \sqrt{2 \times (9 \times 10^{-31} \text{ kg}) \times (1.28 \times 10^{-17} \text{ J})} \]

Calculating inside the square root:

\[ P = \sqrt{2 \times 9 \times 1.28 \times 10^{-48}} = \sqrt{23.04 \times 10^{-48}} = 4.8 \times 10^{-24} \text{ kg m/s} \]

### Step 3: Calculate the de Broglie Wavelength
The de Broglie wavelength \(\lambda\) is given by the formula:

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

Where \(h\) is Planck's constant \(= 6.6 \times 10^{-34} \text{ J s}\).

Substituting the values:

\[ \lambda = \frac{6.6 \times 10^{-34} \text{ J s}}{4.8 \times 10^{-24} \text{ kg m/s}} \]

Calculating the wavelength:

\[ \lambda = 1.375 \times 10^{-10} \text{ m} = 1.375 \text{ Å} \]

### Final Answer
The de Broglie wavelength of the electron is approximately \(1.375 \text{ Å}\).

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