The de broglie wavelength of electron moving with kinetic energy of 144 eV is nearly

Options

A. `102 x 10^(-3) nm`
B. `102 x 10^(-4) nm`
C. `102 x 10^(-5) nm`
D. `102 x 10^(-2) nm`

Answer

To find the de Broglie wavelength of an electron moving with a kinetic energy of 144 eV, we can follow these steps:

### Step 1: Convert Kinetic Energy to Joules
The kinetic energy (KE) is given in electron volts (eV). We need to convert this to joules (J) using the conversion factor:
1 eV = \(1.6 \times 10^{-19}\) J.

\[ KE = 144 \, \text{eV} = 144 \times 1.6 \times 10^{-19} \, \text{J} = 2.304 \times 10^{-17} \, \text{J} \]

### Step 2: Calculate the Momentum of the Electron
The kinetic energy can also be expressed in terms of momentum (p) as follows:

\[ KE = \frac{p^2}{2m} \]

Rearranging this gives us:

\[ p = \sqrt{2m \cdot KE} \]

The mass of the electron (m) is \(9.11 \times 10^{-31}\) kg. Plugging in the values:

\[ p = \sqrt{2 \cdot (9.11 \times 10^{-31} \, \text{kg}) \cdot (2.304 \times 10^{-17} \, \text{J})} \]

Calculating this:

\[ p = \sqrt{4.198 \times 10^{-47}} \approx 6.44 \times 10^{-24} \, \text{kg m/s} \]

### 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, \(h = 6.63 \times 10^{-34} \, \text{J s}\). Substituting the values we have:

\[ \lambda = \frac{6.63 \times 10^{-34} \, \text{J s}}{6.44 \times 10^{-24} \, \text{kg m/s}} \]

Calculating this gives:

\[ \lambda \approx 1.03 \times 10^{-10} \, \text{m} \]

### Step 4: Convert to Picometers
To express this in picometers (pm), we convert meters to picometers:

\[ 1 \, \text{m} = 10^{12} \, \text{pm} \]

Thus,

\[ \lambda \approx 1.03 \times 10^{-10} \, \text{m} = 103 \, \text{pm} \]

### Final Answer
The de Broglie wavelength of the electron moving with a kinetic energy of 144 eV is approximately \(103 \, \text{pm}\).

---