Question

When an electron in an excited molybdenum atom falls from the L to the K shell, an X-ray is emitted. These X-rays are diffracted at an angle of 7.75° by planes with a separation of 2.64 Å. What is the difference in energy between the K shell and the L shell in molybdenum assuming a first order diffraction?

Answer 1

The difference in energy between the K and L shells in molybdenum is approximately 144 eV, and the frequency of the emitted X-ray is approximately 3.48 × 10^16 Hz.

Step-by-step explanation:

The energy difference between the K and L shells in a molybdenum atom can be calculated using the formula ▲EL→K = (Z − 1)² (10.2 eV), where Z is the atomic number. In this case, Z = 13 for molybdenum. So, ▲EL→K = (13 − 1)² (10.2 eV) = 144 eV.

Now, the frequency of the X-ray emitted can be calculated using the formula ƒ = (▲EL→K) / h, where h is Planck's constant. Substituting the values, we get ƒ = 144 eV / 4.14 × 10^-15 eV·s ≈ 3.48 × 10^16 Hz.

Therefore, the difference in energy between the K shell and the L shell in molybdenum is approximately 144 eV and the frequency of the X-ray emitted is approximately 3.48 × 10^16 Hz.