In an electron microscope the electrons generate the
image. For one of the scanning electron microscopes
in our Central Analytical Facility, the electrons are
accelerated to have a kinetic energy of 200,000 eV.
Electrons with this kinetic energy have a velocity of
2.67 x 10^8 m/s, almost 90% of the speed of light. What is the wavelength of the electrons traveling at this very high speed?
I know the right answer is 2.72 pm but I need help getting to that answer
To find the wavelength of the electrons traveling at a high speed, we can use the de Broglie wavelength formula, which relates the wavelength of a particle to its momentum.
The de Broglie wavelength (λ) can be calculated using the following equation:
λ = h / p
Where:
λ is the wavelength
h is the Planck's constant (6.626 x 10^-34 J·s)
p is the momentum of the particle
The momentum (p) can be calculated using the formula:
p = m * v
Where:
m is the mass of the particle
v is the velocity of the particle
In this case, the mass of an electron is approximately 9.10938356 × 10^-31 kg. Given that the electrons are traveling at a velocity of 2.67 x 10^8 m/s, we can calculate the momentum of the electrons:
p = (9.10938356 × 10^-31 kg) * (2.67 x 10^8 m/s)
p ≈ 2.434 x 10^-22 kg·m/s
Now, we can substitute the momentum into the de Broglie wavelength formula:
λ = (6.626 x 10^-34 J·s) / (2.434 x 10^-22 kg·m/s)
λ ≈ 2.72 x 10^-12 m
Since the problem refers to the wavelength in picometers (pm), we can convert the answer:
λ = 2.72 x 10^-12 m * (1 x 10^12 pm / 1 m)
λ ≈ 2.72 pm
Therefore, the wavelength of the electrons traveling at this high speed is approximately 2.72 picometers (pm).