Calculate the wavelength associated with an electron moving with a velocity of `10^ (10)` cm per sec.

Answer

To calculate the wavelength associated with an electron moving with a velocity of \(10^{10}\) cm/s, we will use the de Broglie wavelength formula:

\[ \lambda = \frac{h}{mv} \]

where:
- \(\lambda\) is the wavelength,
- \(h\) is Planck's constant,
- \(m\) is the mass of the electron,
- \(v\) is the velocity of the electron.

### Step-by-Step Solution:

**Step 1: Gather the necessary constants.**
- Mass of the electron, \(m = 9.1 \times 10^{-31}\) kg.
- Planck's constant, \(h = 6.62 \times 10^{-34}\) J·s.
- Velocity of the electron, \(v = 10^{10}\) cm/s.

**Step 2: Convert the mass of the electron to grams.**
Since the velocity is given in cm/s, we can convert the mass of the electron from kg to grams for consistency:
\[ m = 9.1 \times 10^{-31} \text{ kg} = 9.1 \times 10^{-28} \text{ g} \]

**Step 3: Convert Planck's constant to appropriate units.**
We can also convert Planck's constant to the same units as the mass and velocity. However, since we are using \(v\) in cm/s, we will keep \(h\) in J·s, which is fine for our calculation.

**Step 4: Substitute the values into the de Broglie equation.**
Now, we can substitute the values into the de Broglie equation:
\[ \lambda = \frac{h}{mv} = \frac{6.62 \times 10^{-34} \text{ J·s}}{(9.1 \times 10^{-31} \text{ kg}) \times (10^{10} \text{ cm/s})} \]

**Step 5: Calculate the denominator.**
Calculating the denominator:
\[ mv = (9.1 \times 10^{-31} \text{ kg}) \times (10^{10} \text{ cm/s}) = 9.1 \times 10^{-21} \text{ kg·cm/s} \]

**Step 6: Calculate the wavelength.**
Now substituting back into the equation:
\[ \lambda = \frac{6.62 \times 10^{-34}}{9.1 \times 10^{-21}} \text{ cm} \]
Calculating this gives:
\[ \lambda \approx 7.27 \times 10^{-14} \text{ cm} \]

**Step 7: Convert to Angstroms.**
To convert centimeters to angstroms (1 cm = \(10^{10}\) angstroms):
\[ \lambda \approx 7.27 \times 10^{-14} \text{ cm} \times 10^{10} \text{ angstroms/cm} \approx 7.27 \times 10^{-4} \text{ angstroms} \]

### Final Answer:
The wavelength associated with the electron is approximately \(7.27 \times 10^{-4}\) angstroms.