To find the De Broglie wavelength of an electron with a given energy of 100 keV, we can use the De Broglie wavelength formula:

$\lambda = \frac{h}{p}$

where:

• $\lambda$ is the De Broglie wavelength,
• $h$ is Planck's constant ($h \approx 6.626 \times 10^{-34} \, \text{Js}$),
• $p$ is the momentum of the particle.

For an electron with kinetic energy $E$ (100 keV in this case), the momentum $p$ can be derived from the energy using the relation:

$E = \frac{p^2}{2m}$

where $m$ is the mass of the electron ($m \approx 9.11 \times 10^{-31} \, \text{kg}$).

Since $E$ is in keV, let's convert it to joules: $100 \, \text{keV} = 100 \times 10^3 \times 1.602 \times 10^{-19} \, \text{J} = 1.602 \times 10^{-14} \, \text{J}$

Step-by-Step Solution

1. Calculate the momentum $p$: $p = \sqrt{2mE} = \sqrt{2 \times 9.11 \times 10^{-31} \times 1.602 \times 10^{-14}}$

2. Calculate the wavelength $\lambda$: Using $\lambda = \frac{h}{p}$, substitute $h \approx 6.626 \times 10^{-34} \, \text{Js}$.

Let me calculate these values for you.The De Broglie wavelength of an electron with an energy of 100 keV is approximately $3.88 \times 10^{-12}$ meters, or 3.88 picometers.

Would you like more details on the calculations or have further questions?

Here are some related questions to expand on this topic:

1. How does the De Broglie wavelength change with increasing electron energy?
2. What is the significance of the De Broglie wavelength in quantum mechanics?
3. How would the wavelength change if we used a proton instead of an electron?
4. How does the relativistic effect alter the De Broglie wavelength at higher energies?
5. What applications rely on the De Broglie wavelength of particles?

Tip: Remember that as particle energy increases, its De Broglie wavelength decreases, leading to higher resolving power in applications like electron microscopy.