Find the de-Broglie wavelength of an electron with kinetic energy of 120 eV.

Options

A. 112 pm
B. 95 pm
C. 124 pm
D. 102 pm

Answer

To find the de-Broglie wavelength of an electron with a kinetic energy of 120 eV, we can follow these steps:

### Step 1: Understand the de-Broglie wavelength formula
The de-Broglie wavelength (\( \lambda \)) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

### Step 2: Relate kinetic energy to momentum
The kinetic energy (\( KE \)) of a particle can be expressed in terms of momentum:
\[ KE = \frac{p^2}{2m} \]
From this equation, we can solve for momentum \( p \):
\[ p = \sqrt{2m \cdot KE} \]

### Step 3: Substitute kinetic energy in electron volts
For an electron, the kinetic energy can also be expressed as:
\[ KE = e \cdot V \]
where \( e \) is the charge of the electron (approximately \( 1.6 \times 10^{-19} \) coulombs) and \( V \) is the potential difference in volts. Given that the kinetic energy is 120 eV, we have:
\[ KE = 120 \, \text{eV} \]

### Step 4: Convert kinetic energy to joules
To use the kinetic energy in the momentum formula, we need to convert eV to joules. The conversion is:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Thus,
\[ KE = 120 \, \text{eV} = 120 \times 1.6 \times 10^{-19} \, \text{J} = 1.92 \times 10^{-17} \, \text{J} \]

### Step 5: Find the mass of the electron
The mass of the electron (\( m \)) is:
\[ m = 9.1 \times 10^{-31} \, \text{kg} \]

### Step 6: Calculate momentum
Now substitute the values into the momentum formula:
\[ p = \sqrt{2m \cdot KE} = \sqrt{2 \cdot (9.1 \times 10^{-31} \, \text{kg}) \cdot (1.92 \times 10^{-17} \, \text{J})} \]
Calculating this gives:
\[ p = \sqrt{3.4944 \times 10^{-47}} \approx 5.91 \times 10^{-24} \, \text{kg m/s} \]

### Step 7: Calculate the de-Broglie wavelength
Now, substitute \( p \) back into the de-Broglie wavelength formula:
\[ \lambda = \frac{h}{p} \]
where \( h = 6.626 \times 10^{-34} \, \text{J s} \):
\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{J s}}{5.91 \times 10^{-24} \, \text{kg m/s}} \approx 1.12 \times 10^{-10} \, \text{m} \]

### Step 8: Convert to picometers
Since \( 1 \, \text{pm} = 10^{-12} \, \text{m} \):
\[ \lambda \approx 1.12 \times 10^{-10} \, \text{m} = 112 \, \text{pm} \]

### Final Answer
The de-Broglie wavelength of the electron with kinetic energy of 120 eV is approximately **112 pm**.

---