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

Hint: 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 \[\lambda \].