Electrons in a beam incident on a crystal at an angle of 30ͦ have kinetic energies ranging from zero
to a maximum of 5500 eV. The crystal has a grating space d=0.5 Aͦ, and the reflected electrons are
passed through a slit .Find the velocities of the electrons passing through the slit. How many are these
velocities?
To find the velocities of the electrons passing through the slit, we need to use the energy-to-velocity conversion formula for electrons.
The formula for the kinetic energy of an electron can be given as:
KE = (1/2)mv^2
Where:
KE is the kinetic energy,
m is the mass of the electron, and
v is the velocity of the electron.
Since we are given the maximum kinetic energy of the electrons (5500 eV), we can use this information to find the velocity of the electrons.
First, we need to convert the energy from electron volts (eV) to joules (J) since the formula uses SI units. The conversion factor is given as 1 eV = 1.602 x 10^-19 J.
So, 5500 eV = 5500 * 1.602 x 10^-19 J = 8.807 x 10^-16 J.
Now, we can rearrange the kinetic energy formula to solve for the velocity:
v = sqrt(2KE / m)
The mass of an electron, m, is approximately 9.109 x 10^-31 kg.
Plugging in the values, we have:
v = sqrt(2 * 8.807 x 10^-16 J / 9.109 x 10^-31 kg)
Calculating this equation, we get:
v ≈ 2.189 x 10^6 m/s
Therefore, the velocity of the electrons passing through the slit is approximately 2.189 x 10^6 m/s.
As for the number of electrons with these velocities, it is not explicitly given in the question. It would depend on the specifics of the experimental setup and the beam intensity.