The de Broglie wavelength of an electron with kinetic energy 120 e V is `("Given h"=6.63xx10^(-34)Js,m_(e)=9xx10^(-31)kg,1e V=1.6xx10^(-19)J)`

Options

A. `2.13Ã…`
B. `1.13Ã…`
C. `4.15Ã…`
D. `3.14Ã…`

Explanation

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 formula for de Broglie wavelength
The de Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the electron.

### Step 2: Relate momentum to kinetic energy
The momentum \( p \) of an electron can be expressed in terms of its mass \( m \) and kinetic energy \( KE \):
\[ p = \sqrt{2m \cdot KE} \]
Thus, we can rewrite the de Broglie wavelength as:
\[ \lambda = \frac{h}{\sqrt{2m \cdot KE}} \]

### Step 3: Convert kinetic energy from eV to Joules
Given that the kinetic energy \( KE \) is 120 eV, we need to convert this to Joules using the conversion \( 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \):
\[ KE = 120 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 1.92 \times 10^{-17} \text{ J} \]

### Step 4: Substitute values into the formula
Now we can substitute the values into the de Broglie wavelength formula. We have:
- \( h = 6.63 \times 10^{-34} \text{ Js} \)
- \( m = 9 \times 10^{-31} \text{ kg} \)
- \( KE = 1.92 \times 10^{-17} \text{ J} \)

Substituting these values:
\[ \lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \cdot (9 \times 10^{-31}) \cdot (1.92 \times 10^{-17})}} \]

### Step 5: Calculate the denominator
First, calculate the term inside the square root:
\[ 2 \cdot (9 \times 10^{-31}) \cdot (1.92 \times 10^{-17}) = 3.4544 \times 10^{-47} \]
Now take the square root:
\[ \sqrt{3.4544 \times 10^{-47}} \approx 5.88 \times 10^{-24} \]

### Step 6: Calculate the de Broglie wavelength
Now substitute back into the equation for \( \lambda \):
\[ \lambda = \frac{6.63 \times 10^{-34}}{5.88 \times 10^{-24}} \approx 1.13 \times 10^{-10} \text{ m} \]

### Step 7: Convert to Angstroms
Since \( 1 \text{ Angstrom} = 10^{-10} \text{ m} \):
\[ \lambda \approx 1.13 \text{ Angstrom} \]

### Final Answer
The de Broglie wavelength of the electron is approximately \( 1.13 \text{ Angstrom} \).

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