. 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 have the answer i wan the process :2.72 pm
wavelength = h/mv
h = Planck's constant
m = mass electron in kg.
v = velocity in m/s
2.72 pm is correct.
To find the wavelength of electrons traveling at a high speed, we can use the de Broglie wavelength equation:
λ = h / p
where λ is the wavelength, h is Planck's constant (6.626 x 10^-34 J·s), and p is the momentum of the electrons.
The momentum of an object can be calculated using the equation:
p = mv
where p is momentum, m is mass, and v is velocity.
In this case, we are given the kinetic energy of the electrons, but we can use the equation for kinetic energy:
K.E. = (1/2)mv^2
Rearranging the equation to solve for mass (m), we get:
m = (2K.E.) / v^2
Substituting the given values of kinetic energy (200,000 eV) and velocity (2.67 x 10^8 m/s) into the equation, we can calculate the mass of the electrons.
Next, we can substitute the calculated mass and velocity into the equation for momentum to find the momentum of the electrons.
Finally, we can substitute the calculated momentum and Planck's constant into the de Broglie wavelength equation to find the wavelength of the electrons.
Using this process, we find that the wavelength of the electrons traveling at this high speed is 2.72 picometers (pm).