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).