An electron experiment was performed with a beam of electron accelerated by a potential difference of `10.0 k eV` .What is the wavelength of the electron beam

Explanation

To find the wavelength of an electron beam accelerated by a potential difference of 10.0 k eV, we can follow these steps:

### Step 1: Calculate the Kinetic Energy of the Electron
The kinetic energy (KE) gained by an electron when it is accelerated through a potential difference (V) is given by the formula:
\[ KE = eV \]
where:
- \( e \) is the charge of the electron, approximately \( 1.602 \times 10^{-19} \) Coulombs.
- \( V \) is the potential difference in volts (10.0 kV = \( 10.0 \times 10^3 \) V).

Substituting the values:
\[ KE = (1.602 \times 10^{-19} \, \text{C}) \times (10.0 \times 10^3 \, \text{V}) \]
\[ KE = 1.602 \times 10^{-16} \, \text{J} \]

### Step 2: Calculate the Velocity of the Electron
The kinetic energy can also be expressed in terms of mass and velocity:
\[ KE = \frac{1}{2} mv^2 \]
Rearranging this gives:
\[ v = \sqrt{\frac{2KE}{m}} \]
where \( m \) is the mass of the electron, approximately \( 9.1 \times 10^{-31} \) kg.

Substituting the values:
\[ v = \sqrt{\frac{2 \times 1.602 \times 10^{-16} \, \text{J}}{9.1 \times 10^{-31} \, \text{kg}}} \]
Calculating this gives:
\[ v \approx 5.93 \times 10^7 \, \text{m/s} \]

### Step 3: Calculate the Wavelength of the Electron Beam
The wavelength \( \lambda \) of the electron can be calculated using the de Broglie wavelength formula:
\[ \lambda = \frac{h}{mv} \]
where \( h \) is Planck's constant, approximately \( 6.626 \times 10^{-34} \, \text{J s} \).

Substituting the values:
\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{J s}}{(9.1 \times 10^{-31} \, \text{kg}) \times (5.93 \times 10^7 \, \text{m/s})} \]
Calculating this gives:
\[ \lambda \approx 1.23 \times 10^{-8} \, \text{m} \]
This can be converted to angstroms:
\[ \lambda \approx 123 \, \text{Å} \]

### Final Answer
The wavelength of the electron beam is approximately **123 angstroms**.

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