Question

The wavelength of characteristic K\(_\alpha\) X-ray photons from Mo (atomic number 42) is ............. Å. (Round off to one decimal place).

Hint:

The characteristic X-ray wavelengths can be calculated using the Rydberg formula, where \( Z \) is the atomic number, and \( n_1 \) and \( n_2 \) are the principal quantum numbers of the initial and final energy levels.

Answer 1

Correct Answer: 0.6

Step 1: Use the Rydberg formula for X-ray wavelengths.
The characteristic K\(_\alpha\) X-ray wavelength for an atom can be calculated using the formula: \[ \lambda = \frac{R \cdot Z^2}{n_1^2 - n_2^2}, \] where \[ R = 1.097 \times 10^7 \, \text{m}^{-1} \quad \text{(Rydberg constant)} \quad \text{and} \quad Z = 42 \quad \text{(atomic number of Mo)}. \] For the K\(_\alpha\) transition, \( n_1 = 2 \) and \( n_2 = 1 \), so \[ \lambda = \frac{1.097 \times 10^7 \times 42^2}{2^2 - 1^2}. \] Step 2: Simplify and calculate the wavelength.
First, calculate the numerator: \[ 42^2 = 1764, \quad 1.097 \times 10^7 \times 1764 = 1.938 \times 10^{10}. \] Now, calculate the denominator: \[ 2^2 - 1^2 = 4 - 1 = 3. \] Thus, \[ \lambda = \frac{1.938 \times 10^{10}}{3} = 6.46 \times 10^9 \, \text{m}. \] Converting to nanometers: \[ \lambda = 6.46 \, \text{nm} = 0.646 \, \text{Å}. \] Final Answer: The wavelength of K\(_\alpha\) X-ray photons from Mo is \( \boxed{0.6} \, \text{Å}. \)