What is the kinetic energy of each electron in a beam of electrons if the beam produces a diffraction pattern of a crystal which is similar to that of a beam of 1.00 eV neutrons? Note: The electron mass is 9.11 × 10–31kg; and the neutron mass is 1.67 × 10–27 kg.

I know the answer is 1830 eV, but I have no idea how to get this answer.

Can someone please help? Can you provide a thorough explanation, step-by-step, not just the answer.

Thanks so much!

To determine the kinetic energy of each electron in the beam, you need to follow a step-by-step process using the given information and relevant formulas.

Step 1: Understand the concept
The concept here is that the diffraction pattern produced by the beam of electrons is similar to that of a beam of 1.00 eV neutrons. The energy of the neutrons is given, and you need to determine the kinetic energy of the electrons.

Step 2: Kinetic energy formula
The formula for kinetic energy is given by:

KE = (1/2)mv^2

Where KE is the kinetic energy, m is the mass of the particle, and v is its velocity.

Step 3: Compare mass of electrons and neutrons
Given:
Mass of an electron (m_e) = 9.11 × 10^(-31) kg
Mass of a neutron (m_n) = 1.67 × 10^(-27) kg

Clearly, the mass of electrons is much smaller (around 2000 times) than the mass of neutrons.

Step 4: Relationship between velocity and kinetic energy
Since we are given the energy of the neutrons (1.00 eV), we can convert energy to kinetic energy using the relationship: E = (1/2)mv^2.
So, v^2 = (2E)/m.

Step 5: Calculate the velocity of the neutrons
Given:
Energy of the neutrons (E_n) = 1.00 eV.
Convert it to Joules: 1 eV = 1.6 × 10^(-19) J.
Substituting the values, we get:
E_n = (1/2)m_n(v_n^2).
(v_n^2) = (2E_n)/m_n.
(v_n^2) = (2 × 1.6 × 10^(-19) J) / (1.67 × 10^(-27) kg).
Solve for (v_n^2).

Step 6: Calculate the velocity of the electrons
Now, we know that the diffraction pattern produced by the electrons is similar to that of the neutrons. Therefore, the electrons and neutrons must have the same velocity.

Step 7: Calculate the kinetic energy of the electrons
Finally, we need to calculate the kinetic energy of the electrons.
Given:
Mass of electrons (m_e) = 9.11 × 10^(-31) kg.
Velocity of electrons (v_e) = (v_n) (from step 6, as they have the same velocity).

Substituting the known values, we can calculate the kinetic energy of electrons as follows:
KE_e = (1/2)m_e(v_e^2).

Step 8: Convert the kinetic energy to electron volt
The kinetic energy in joules can be converted to electron volts (eV) by using the conversion factor:
1 eV = 1.6 × 10^(-19) J.

Performing the calculations as described above, you will find that the kinetic energy of each electron in the beam is approximately 1830 eV.