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 relationship between kinetic energy and velocity. The kinetic energy of an electron can be calculated using the equation:
K.E. = (1/2) * m * v^2,
where K.E. is the kinetic energy, m is the mass of the electron, and v is the velocity of the electron.
In this case, we are given the maximum kinetic energy of the electrons, which is 5500 eV. We also know that the electrons are incident on the crystal at an angle of 30 degrees. This implies that the kinetic energy of the electrons will vary from zero to the maximum value.
We can use the maximum kinetic energy to find the maximum velocity of the electrons. Rearranging the equation above, we get:
v^2 = (2 * K.E.) / m.
Substituting the values into the equation, we have:
v^2 = (2 * 5500 eV) / m.
Since m is the mass of the electron, we can use the mass-energy equivalence, E = mc^2, to convert the electron mass into energy. The rest mass of an electron is approximately 0.511 MeV/c^2. Therefore, we can write:
m = (0.511 MeV/c^2) / (c^2),
m = 0.511 MeV / c^2.
Now we can substitute the values of m and K.E. into the equation for v^2:
v^2 = (2 * 5500 eV) / (0.511 MeV / c^2).
To simplify the units, we can convert eV to MeV by dividing by 1 million:
v^2 = (2 * 5500) / (0.511).
Simplifying further, we get:
v^2 ≈ 21575.449.
Taking the square root of both sides gives us the maximum velocity:
v ≈ √(21575.449).
Therefore, the maximum velocity of the electrons passing through the slit is approximately √(21575.449) m/s.
To determine how many electrons have this velocity, we need more information about the incident electron beam. Specifically, we would need to know the number of electrons per unit area or per second in the beam. Without that information, we cannot determine the exact number of electrons with a specific velocity.