Wavelength of an electron is 5Å. Velocity of the electron is

Options

A. `1.45 xx 10^(8) cm//s`
B. `1.6 xx 10^(-8) cm//s`
C. `3.2 xx 10^(-27) cm//s`
D. `3.2 xx 10^(27) cm//s`

Answer

To find the velocity of an electron given its wavelength, we can use the de Broglie wavelength formula, which relates the wavelength (λ) of a particle to its momentum (p). The formula is:

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

Where:
- \( \lambda \) is the wavelength,
- \( h \) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Joule second}\)),
- \( p \) is the momentum of the electron.

The momentum \( p \) can also be expressed as:

\[ p = m \cdot v \]

Where:
- \( m \) is the mass of the electron (\(9.1 \times 10^{-31} \, \text{kg}\)),
- \( v \) is the velocity of the electron.

### Step-by-step Solution:

1. **Identify the given values:**
- Wavelength \( \lambda = 5 \, \text{Å} = 5 \times 10^{-10} \, \text{m} \)
- Planck's constant \( h = 6.626 \times 10^{-34} \, \text{Joule second} \)
- Mass of the electron \( m = 9.1 \times 10^{-31} \, \text{kg} \)

2. **Rearrange the de Broglie equation to solve for velocity \( v \):**
\[ v = \frac{h}{\lambda \cdot m} \]

3. **Substitute the known values into the equation:**
\[ v = \frac{6.626 \times 10^{-34}}{(5 \times 10^{-10}) \cdot (9.1 \times 10^{-31})} \]

4. **Calculate the denominator:**
\[ (5 \times 10^{-10}) \cdot (9.1 \times 10^{-31}) = 4.55 \times 10^{-40} \]

5. **Now substitute back into the equation for \( v \):**
\[ v = \frac{6.626 \times 10^{-34}}{4.55 \times 10^{-40}} \]

6. **Perform the division:**
\[ v \approx 1.46 \times 10^{6} \, \text{m/s} \]

7. **Convert the velocity to cm/s:**
\[ v \approx 1.46 \times 10^{6} \, \text{m/s} = 1.46 \times 10^{8} \, \text{cm/s} \]

### Final Answer:
The velocity of the electron is approximately \( 1.46 \times 10^{8} \, \text{cm/s} \).