Question

The de Broglie wavelength of an electron that has been accelerate through a potential difference of 100 V

• A. 1.8nm
• B. 13.2nm
• C. 0.123nm
• D. 1.23.nm

Answer 1

The Correct Option is C

To calculate the de Broglie wavelength of an electron accelerated through a potential difference of 100 V, we'll follow these steps:

1. Given Data:
- Potential difference (V) = 100 V
- Charge of electron (e) = 1.602 × 10^-19 C
- Mass of electron (m) = 9.109 × 10^-31 kg
- Planck's constant (h) = 6.626 × 10^-34 J·s

2. Calculate the Kinetic Energy:
The kinetic energy gained by the electron is:
KE = eV = (1.602 × 10^-19 C)(100 V) = 1.602 × 10^-17 J

3. Determine the Electron's Velocity:
Using KE = ½mv²:
v = √(2KE/m) = √(2 × 1.602 × 10^-17 J / 9.109 × 10^-31 kg)
≈ 5.93 × 10⁶ m/s

4. Calculate the de Broglie Wavelength:
Using λ = h/p = h/(mv):
λ = (6.626 × 10^-34 J·s) / (9.109 × 10^-31 kg × 5.93 × 10⁶ m/s)
≈ 1.23 × 10^-10 m = 1.23 Å

5. Simplified Formula (Alternative Approach):
For electrons, we can use the shortcut formula:
λ = h/√(2meV)
= 6.626 × 10^-34/√(2 × 9.109 × 10^-31 × 1.602 × 10^-19 × 100)
= 1.23 × 10^-10 m

6. Interpretation:
This wavelength (1.23 Å) is comparable to atomic spacing in crystals, which is why electron beams are useful for studying crystal structures.

Final Answer:
The de Broglie wavelength of an electron accelerated through 100 V is 1.23 × 10^-10 m (or 1.23 Å).