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 calculate the velocity of each electron using its kinetic energy. The formula to calculate the velocity of an electron is as follows:
v = √((2eV) / m)
where v is the velocity of the electron, e is the charge of the electron (1.6 x 10^-19 C), V is the kinetic energy of the electron (in joules), and m is the mass of the electron (9.109 x 10^-31 kg).
Given that the maximum kinetic energy of the electrons is 5500 eV, we need to convert it to joules by multiplying it by the conversion factor of 1.6 x 10^-19 J/eV. Hence,
V = 5500 eV * (1.6 x 10^-19 J/eV) = 8.8 x 10^-16 J
Now, we can substitute the values into the formula to calculate the velocity of the electrons:
v = √((2 * 1.6 x 10^-19 C * 8.8 x 10^-16 J) / 9.109 x 10^-31 kg)
Simplifying the equation:
v = √(2.816 x 10^-34 C.J / 9.109 x 10^-31 kg)
v = √(3.09 x 10^3 m^2/s^2)
v ≈ 55.6 m/s
Therefore, the velocity of the electrons passing through the slit is approximately 55.6 m/s.
To determine how many electrons have this velocity, we need to consider the distribution of electron energies. Since we know the range of kinetic energies from zero to a maximum of 5500 eV, we can assume that the distribution is continuous and follows a probability density function (PDF).
To calculate the number of electrons with a particular velocity, we integrate the PDF over the range of velocities corresponding to that particular velocity. However, without additional information about the PDF, it is not possible to determine the exact number of electrons with a velocity of 55.6 m/s.