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 can use the formula relating kinetic energy to velocity for non-relativistic particles:

E = (1/2) mv^2

Where E is the kinetic energy, m is the mass of the electron, and v is the velocity.

First, convert the maximum kinetic energy to joules since the unit "eV" is electron volts:

5500 eV * (1.602 x 10^-19 J/eV) = 8.812 x 10^-16 J

The mass of an electron, m, is approximately 9.11 x 10^-31 kg.

Now, rearrange the formula to solve for the velocity:

v = √(2E / m)

Substituting the values:

v = √(2 * 8.812 x 10^-16 J / 9.11 x 10^-31 kg)

Calculating the square root:

v ≈ 2.285 x 10^7 m/s

So, the velocity of the electrons passing through the slit is approximately 2.285 x 10^7 m/s.

To calculate the number of velocities, we need to determine the number of different kinetic energies corresponding to the range from zero to the maximum energy.

Divide the maximum kinetic energy by the smallest increment of energy to find the number of velocities:

Number of velocities = (Maximum kinetic energy) / (Smallest increment of energy)

Since the kinetic energies range from zero to the maximum of 5500 eV, the smallest increment of energy is 1 eV.

Number of velocities = 5500 eV / 1 eV = 5500

So, there are 5500 different velocities for the electrons passing through the slit.