Question

What will be the wavelength of ${{\text{K}}_{\alpha}}$ line for $z{\text{ = 31}}$ when $\alpha = 5 \times {10^7}{\text{ Hz}}$ for a characteristic X-ray spectrum?
A. $1.33\,\mathop {\text{A}}\limits^{\text{o}}$
B. $1.33\,{\text{nm}}$
C. $133 \times {10^{ - 10}}\,{\text{m}}$
D. $133\,{\text{nm}}$

Answer 1

Use the formula $\dfrac{1}{\lambda } = R{\left( {z - 1} \right)^2} \times \left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$, take ${n_1}{\text{ = 1}}$ and ${n_2}{\text{ = 2}}$.

Complete step by step solution:
Characteristic X-rays are released when electrons from the outer shell fill a gap in an atom's inner shell, releasing X-rays in a pattern that is "characteristic" for each particle. Characteristic X-rays are created when an element is bombarded with particles of high energy that may be photons, electrons or ions (such as protons). If a bound electron (the target electron) in an atom is struck by the incident particle, the target electron is expelled from the atom's inner shell. Each element has a particular set of levels of energy, and thus the transition from higher to lower levels of energy generates X-rays with frequencies characteristic of each element.
Here, in this case, ${{\text{K}}_{\alpha}}$ line is produced, which defines transition of the x-rays emitted by transitions from the levels $n{\text{ = 2}}$ to $n{\text{ = 1}}$ are called K-alpha x-rays and are called K-beta x-rays for the transition $n{\text{ = 3}}$ to $n{\text{ = 1}}$.

The formula which relates wavelength and transition is:

$\dfrac{1}{\lambda } = R{\left( {z - 1} \right)^2} \times \left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$ …… (1)
Where,
$\lambda$ indicates wavelength.
$R$ indicates Rydberg’s constant.
$z$ indicates atomic number.
${n_1}$ indicates lower energy level.
${n_2}$ indicates higher energy level.

Substituting, the values $R = 1.097 \times {10^7}\,{{\text{m}}^{ - 1}}$, $z{\text{ = 31}}$, ${n_1}{\text{ = 1}}$ and ${n_2}{\text{ = 2}}$ in equation (1):

$$\dfrac{1}{\lambda } = R{\left( {z - 1} \right)^2} \times \left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right) \\\ = 1.097 \times {10^7} \times {\left( {31 - 1} \right)^2} \times \left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}} \right) \\\ = 1.097 \times {10^7} \times {30^2} \times \left( {1 - \dfrac{1}{4}} \right) \\\ = \,1.097 \times {10^7} \times 900 \times \dfrac{3}{4} \\\$$ $$= 1.097 \times {10^7} \times 9 \times {10^2} \times \dfrac{3}{4} \\\ = 7.40 \times {10^9}\,{{\text{m}}^{ - {\text{1}}}} \\\$$

Hence,

$$\dfrac{1}{\lambda } = 7.40 \times {10^9}\,{{\text{m}}^{ - {\text{1}}}} \\\ \lambda = \dfrac{1}{{7.40 \times {{10}^9}}}\,\,{\text{m}} \\\ {\text{ = 0}}{\text{.133}} \times {10^{ - 9}} \\\ = 1.33 \times {10^{ - 10}}\,{\text{m}} \\\ {\text{ = }}1.33\,\,\mathop {\text{A}}\limits^{\text{o}} \\\$$

**Hence, the wavelength is $1.33\,\,\mathop {\text{A}}\limits^{\text{o}}$.

The correct answer is (A).**

Note: In this problem, you are asked to find the wavelength for the ${{\text{K}}_{\alpha}}$ line produced. For this, always remember that the alpha X-rays are produced for the transition from levels $n{\text{ = 2}}$ to $n{\text{ = 1}}$ and beta X-rays are produced for the transition levels $n{\text{ = 3}}$ to $n{\text{ = 1}}$. Don’t be confused at this point. First you will find $\dfrac{1}{\lambda }$ and then find the reverse to find $\lambda$.