In 1924, de-Broglie proposed that every particle possesses wave properties with a wavelength , `lambda` given by `lambda = h/(mv)` where m is the mass of the particle, `v` is its velocity and h is Planck's constant. The de-Broglie prediction was confirmed experimentally when it was found that an electron beam undergoes diffraction , a phenomenon characteristic of waves. The de-Broglie wavelength can be estimated by measuring kinetic energy of an electron accelerating by a potential V as :
`1/2 mv^2 = eV "where" 1eV = 1.6 xx 10^(-19) J, h = 6.6 xx 10^(-34) Js`.
The wavelength of matter wave associated with an electron passing through an electric potential of 100 million volts is

Options

A. `10^2 (2 m e)^(1//2)h`
B. `(h xx 10^(-4))/((2 m e)^(1//2))`
C. `(10^(-3)h)/((2 m e)^(1//2))`
D. `(h xx 10^(-4))/((2 m e)^(-1//2))`

Explanation

To find the wavelength of the matter wave associated with an electron passing through an electric potential of 100 million volts (100 MV), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Electric potential, \( V = 100 \text{ MV} = 100 \times 10^6 \text{ V} = 1 \times 10^8 \text{ V} \)
- Planck's constant, \( h = 6.6 \times 10^{-34} \text{ Js} \)
- Charge of an electron, \( e = 1.6 \times 10^{-19} \text{ C} \)

2. **Calculate the Kinetic Energy of the Electron:**
The kinetic energy \( KE \) of an electron accelerated through a potential \( V \) is given by:
\[ KE = eV \]
Substituting the values:
\[ KE = (1.6 \times 10^{-19} \text{ C})(1 \times 10^8 \text{ V}) = 1.6 \times 10^{-11} \text{ J} \]

3. **Relate Kinetic Energy to Velocity:**
The kinetic energy is also expressed as:
\[ KE = \frac{1}{2} mv^2 \]
Rearranging for \( v \):
\[ v = \sqrt{\frac{2 \cdot KE}{m}} \]
The mass of an electron \( m \) is approximately \( 9.11 \times 10^{-31} \text{ kg} \). Substituting the values:
\[ v = \sqrt{\frac{2 \cdot (1.6 \times 10^{-11})}{9.11 \times 10^{-31}}} \]

4. **Calculate the Velocity \( v \):**
\[ v = \sqrt{\frac{3.2 \times 10^{-11}}{9.11 \times 10^{-31}}} \approx \sqrt{3.51 \times 10^{19}} \approx 5.93 \times 10^9 \text{ m/s} \]

5. **Calculate the de Broglie Wavelength \( \lambda \):**
The de Broglie wavelength is given by:
\[ \lambda = \frac{h}{mv} \]
Substituting the values:
\[ \lambda = \frac{6.6 \times 10^{-34}}{(9.11 \times 10^{-31})(5.93 \times 10^9)} \]

6. **Perform the Calculation:**
\[ \lambda = \frac{6.6 \times 10^{-34}}{5.39 \times 10^{-21}} \approx 1.22 \times 10^{-13} \text{ m} \]

7. **Convert to Picometers:**
\[ \lambda \approx 1.22 \times 10^{-13} \text{ m} = 122 \text{ pm} \]

### Final Answer:
The wavelength of the matter wave associated with an electron passing through an electric potential of 100 million volts is approximately **122 picometers (pm)**.