Electrons in a beam incident on a crystal at an angle of 300 have kinetic energies ranging from zero
to a maximum of 5500 eV. The crystal has a grating space d=0.5A0, 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 first need to convert the given maximum kinetic energy of 5500 eV into electron volts (eV) to the SI unit of energy, which is joules (J).
1 eV is equivalent to 1.6 x 10^-19 J. Therefore, 5500 eV can be calculated as:
5500 eV * (1.6 x 10^-19 J / 1 eV) = 8.8 x 10^-16 J
Next, we can use the kinetic energy formula to find the velocity of an electron:
Kinetic Energy (KE) = (1/2) * mass * velocity^2
The mass of an electron is approximately 9.11 x 10^-31 kg. Rearranging the formula, we can solve for the velocity:
velocity = sqrt((2 * KE) / mass)
Now, substitute the calculated kinetic energy and electron mass values into the formula:
velocity = sqrt((2 * 8.8 x 10^-16 J) / (9.11 x 10^-31 kg))
Calculate the velocity using a scientific calculator or software, and you will get the velocity of the electrons passing through the slit.
To find how many velocities are there, we need more information about the slit. If the slit width is given, we can calculate the number of velocities based on the range of angles the electrons are incident on the crystal. Please provide information about the slit width if available.