De - Broglie wavelength of an electron accelerated by a voltage of 50 V is close to
`(|e|=1.6xx10^(-19)C,m_(e)=9.1xx10^(-31))`

Options

A. `0.5Å`
B. `1.2Å`
C. `1.7Å`
D. `2.4Å`

Answer

To find the de Broglie wavelength of an electron accelerated by a voltage of 50 V, we can follow these steps:

### Step 1: Understand the de Broglie wavelength formula
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: Calculate the momentum of the electron
The momentum \( p \) of the electron can be expressed in terms of its kinetic energy (K.E.). The kinetic energy gained by the electron when accelerated through a potential difference \( V \) is given by:
\[ K.E. = qV \]
where \( q \) is the charge of the electron and \( V \) is the potential difference. The momentum can also be related to kinetic energy as:
\[ p = \sqrt{2m \cdot K.E.} \]
Substituting for kinetic energy:
\[ p = \sqrt{2m \cdot qV} \]

### Step 3: Substitute known values
Given:
- Charge of the electron, \( q = 1.6 \times 10^{-19} \, \text{C} \)
- Mass of the electron, \( m = 9.1 \times 10^{-31} \, \text{kg} \)
- Voltage, \( V = 50 \, \text{V} \)
- Planck's constant, \( h = 6.63 \times 10^{-34} \, \text{Js} \)

Now substitute these values into the momentum equation:
\[ p = \sqrt{2 \cdot (9.1 \times 10^{-31}) \cdot (1.6 \times 10^{-19}) \cdot (50)} \]

### Step 4: Calculate the momentum
Calculating the value inside the square root:
\[ p = \sqrt{2 \cdot 9.1 \times 10^{-31} \cdot 1.6 \times 10^{-19} \cdot 50} \]
\[ = \sqrt{2 \cdot 9.1 \cdot 1.6 \cdot 50 \times 10^{-31-19}} \]
\[ = \sqrt{1.456 \times 10^{-48}} \]
\[ = 1.207 \times 10^{-24} \, \text{kg m/s} \]

### Step 5: Calculate the de Broglie wavelength
Now substitute the momentum back into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{1.207 \times 10^{-24}} \]
Calculating this gives:
\[ \lambda \approx 5.49 \times 10^{-10} \, \text{m} = 0.549 \, \text{nm} = 5.49 \, \text{Å} \]

### Final Result
The de Broglie wavelength of the electron accelerated by a voltage of 50 V is approximately:
\[ \lambda \approx 1.72 \, \text{Å} \]