To find the wavelength of electrons in an electron beam lithography system with an electron gun of energy 50 keV, we can use the de Broglie wavelength formula:

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

where:

• $\lambda$ is the wavelength
• $h$ is Planck's constant
• $p$ is the momentum of the electron

First, we need to convert the electron energy from keV to joules. The energy given is 50 keV:

$E = 50 \, \text{keV} = 50 \times 10^3 \, \text{eV}$

Since 1 electron volt (eV) is equivalent to $1.602 \times 10^{-19}$ joules (J):

$E = 50 \times 10^3 \times 1.602 \times 10^{-19} \, \text{J}$ $E = 8.01 \times 10^{-15} \, \text{J}$

Next, we use the relationship between energy and momentum for non-relativistic electrons:

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

Solving for momentum $p$:

$p = \sqrt{2mE}$

Substituting the given values:

$m = 9.109 \times 10^{-31} \, \text{kg}$ $E = 8.01 \times 10^{-15} \, \text{J}$

$p = \sqrt{2 \times 9.109 \times 10^{-31} \, \text{kg} \times 8.01 \times 10^{-15} \, \text{J}}$ $p = \sqrt{1.459 \times 10^{-44} \, \text{kg}^2 \, \text{m}^2 / \text{s}^2}$ $p = 1.209 \times 10^{-22} \, \text{kg} \, \text{m/s}$

Now, we can find the wavelength $\lambda$:

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

$h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s}$

$\lambda = \frac{6.626 \times 10^{-34} \, \text{J} \cdot \text{s}}{1.209 \times 10^{-22} \, \text{kg} \, \text{m/s}}$ $\lambda = 5.48 \times 10^{-12} \, \text{m}$

So, the wavelength of the electrons in the electron beam lithography system with a 50 keV electron gun is approximately $5.48 \, \text{pm}$ (picometers).

Would you like more details or have any questions?

Here are some related questions for further exploration:

1. How does electron wavelength affect resolution in electron beam lithography?
2. What is the principle of de Broglie wavelength and its applications in modern technology?
3. How do relativistic effects alter the calculations of electron wavelengths?
4. What are the advantages of electron beam lithography compared to photolithography?
5. How do electron lenses work in focusing the electron beam in lithography systems?
6. What are the limitations of electron beam lithography?
7. How does energy variation of electron beams impact feature size in lithography?
8. What materials are commonly used as resists in electron beam lithography?

Tip: In electron beam lithography, maintaining a vacuum environment is crucial to prevent scattering of electrons by air molecules, ensuring high-resolution patterning.