An electron is accelerated by a potential difference of 50 V. Find the de Broglie wavelength associated with it.

Options

A. `2.7xx10^(-10)`
B. `1.74xx10=^(-10)`
C. `3.6xx10^(-9)`
D. `4.9xx10^(-11)`

Explanation

To find the de Broglie wavelength associated with an electron that has been accelerated by a potential difference of 50 V, we can follow these steps:

### Step 1: Calculate the Kinetic Energy of the Electron
When an electron is accelerated through a potential difference \( V \), it gains kinetic energy given by the equation:
\[ KE = eV \]
where:
- \( e \) is the charge of the electron (\( 1.6 \times 10^{-19} \) C)
- \( V \) is the potential difference (50 V)

Substituting the values:
\[ KE = (1.6 \times 10^{-19} \, \text{C})(50 \, \text{V}) = 8.0 \times 10^{-18} \, \text{J} \]

### Step 2: Relate Kinetic Energy to Velocity
The kinetic energy can also be expressed in terms of mass and velocity:
\[ KE = \frac{1}{2} mv^2 \]
Setting the two expressions for kinetic energy equal gives:
\[ \frac{1}{2} mv^2 = eV \]
Rearranging for \( v \):
\[ v = \sqrt{\frac{2eV}{m}} \]

### Step 3: Substitute Values to Find Velocity
Using the mass of the electron \( m = 9.11 \times 10^{-31} \, \text{kg} \):
\[ v = \sqrt{\frac{2 \times (1.6 \times 10^{-19} \, \text{C}) \times (50 \, \text{V})}{9.11 \times 10^{-31} \, \text{kg}}} \]
Calculating this gives:
\[ v = \sqrt{\frac{8.0 \times 10^{-18}}{9.11 \times 10^{-31}}} \approx 4.19 \times 10^{6} \, \text{m/s} \]

### Step 4: Calculate the Momentum of the Electron
The momentum \( p \) of the electron can be calculated using:
\[ p = mv \]
Substituting the values:
\[ p = (9.11 \times 10^{-31} \, \text{kg})(4.19 \times 10^{6} \, \text{m/s}) \approx 3.81 \times 10^{-24} \, \text{kg m/s} \]

### Step 5: Calculate the de Broglie Wavelength
The de Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{Js} \)). Substituting the values:
\[ \lambda = \frac{6.63 \times 10^{-34}}{3.81 \times 10^{-24}} \approx 1.74 \times 10^{-10} \, \text{m} \]

### Final Result
Thus, the de Broglie wavelength associated with the electron is approximately:
\[ \lambda \approx 1.74 \times 10^{-10} \, \text{m} \]