An electron is accelerated by a potential difference of 9.375 V . Find the de-Broglie wavelength associated with it is Å

Answer

To find the de-Broglie wavelength associated with an electron accelerated by a potential difference of 9.375 V, we can follow these steps:

### Step 1: Understand the formula for de-Broglie wavelength
The de-Broglie wavelength (\( \lambda \)) is given by the formula:
\[ \lambda = \frac{h}{\sqrt{2m e V}} \]
where:
- \( h \) = Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \))
- \( m \) = mass of the electron (\( 9.1 \times 10^{-31} \, \text{kg} \))
- \( e \) = charge of the electron (\( 1.6 \times 10^{-19} \, \text{C} \))
- \( V \) = potential difference (9.375 V)

### Step 2: Substitute the known values into the formula
Substituting the known values into the formula:
\[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times (9.1 \times 10^{-31}) \times (1.6 \times 10^{-19}) \times (9.375)}} \]

### Step 3: Calculate the denominator
First, calculate the product in the denominator:
\[ 2 \times (9.1 \times 10^{-31}) \times (1.6 \times 10^{-19}) \times (9.375) \]
Calculating this step-by-step:
1. Calculate \( 2 \times 9.1 \times 10^{-31} = 1.82 \times 10^{-30} \)
2. Calculate \( 1.82 \times 10^{-30} \times 1.6 \times 10^{-19} = 2.912 \times 10^{-49} \)
3. Calculate \( 2.912 \times 10^{-49} \times 9.375 = 2.726 \times 10^{-48} \)

Now take the square root:
\[ \sqrt{2.726 \times 10^{-48}} \approx 5.22 \times 10^{-24} \]

### Step 4: Calculate the de-Broglie wavelength
Now substitute back into the equation for \( \lambda \):
\[ \lambda = \frac{6.626 \times 10^{-34}}{5.22 \times 10^{-24}} \approx 1.27 \times 10^{-10} \, \text{m} \]

### Step 5: Convert to Angstroms
To convert meters to Angstroms (1 Å = \( 10^{-10} \, \text{m} \)):
\[ \lambda \approx 1.27 \times 10^{-10} \, \text{m} = 1.27 \, \text{Å} \]

### Final Answer:
The de-Broglie wavelength associated with the electron is approximately \( 1.27 \, \text{Å} \).

---