What is de-Broglie wavelength associated with an `e^(-)` accelerated through `P.D=100kV`?

Explanation

To find the de-Broglie wavelength associated with an electron accelerated through a potential difference of 100 kV, 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 electron.

### Step 2: Calculate the momentum of the electron
The momentum \( p \) of the electron can be expressed in terms of its kinetic energy. When an electron is accelerated through a potential difference \( V \), its kinetic energy (\( KE \)) is given by:
\[ KE = eV \]
where \( e \) is the charge of the electron (\( 1.6 \times 10^{-19} \) C) and \( V \) is the potential difference in volts. For \( V = 100 \) kV, we convert it to volts:
\[ V = 100 \times 10^3 \, \text{V} = 10^5 \, \text{V} \]
Thus, the kinetic energy becomes:
\[ KE = eV = (1.6 \times 10^{-19} \, \text{C})(10^5 \, \text{V}) = 1.6 \times 10^{-14} \, \text{J} \]

### Step 3: Relate kinetic energy to momentum
The kinetic energy can also be expressed in terms of momentum:
\[ KE = \frac{p^2}{2m} \]
where \( m \) is the mass of the electron (\( 9.1 \times 10^{-31} \, \text{kg} \)). Rearranging this gives:
\[ p = \sqrt{2m \cdot KE} \]

### Step 4: Substitute the values
Substituting the values of \( m \) and \( KE \):
\[ p = \sqrt{2 \cdot (9.1 \times 10^{-31} \, \text{kg}) \cdot (1.6 \times 10^{-14} \, \text{J})} \]

### Step 5: Calculate the momentum
Calculating the above expression:
\[ p = \sqrt{2 \cdot 9.1 \times 10^{-31} \cdot 1.6 \times 10^{-14}} = \sqrt{2.912 \times 10^{-44}} \approx 5.39 \times 10^{-22} \, \text{kg m/s} \]

### Step 6: Calculate the de-Broglie wavelength
Now, substituting \( p \) back into the de-Broglie wavelength formula:
\[ \lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34} \, \text{J s}}{5.39 \times 10^{-22} \, \text{kg m/s}} \approx 1.23 \times 10^{-12} \, \text{m} \]
Converting this to picometers:
\[ \lambda \approx 1.23 \, \text{pm} \, (\text{pico meters}) \]

### Final Answer
The de-Broglie wavelength associated with an electron accelerated through a potential difference of 100 kV is approximately \( 1.23 \, \text{pm} \).

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