Question

X-ray from a tube with a target $A$ of the atomic number $Z$ shows strong $K$ lines for target $A$ and weak $K$ lines for impurities. The wavelength of ${K_\alpha }$ lines is ${\lambda _z}$ for target $A$ and ${\lambda _1}$ and ${\lambda _2}$ for two impurities.
$\dfrac{{{\lambda _z}}}{{{\lambda _1}}} = 4$ and $\dfrac{{{\lambda _z}}}{{{\lambda _2}}} = \dfrac{1}{4}$. Assuming the screening constant of ${K_\alpha }$ lines to be unity, select the correct statements(s).
A. The atomic number of first impurity is $2z - 1$
B. The atomic number of first impurity is $2z + 1$
C. The atomic number of second impurity is $\dfrac{{\left( {z + 1} \right)}}{2}$
D. The atomic number of second impurity is $\dfrac{z}{2} + 1$

Answer 1

Hint: Moseley’s law of equation is used to carry out this problem. By using the given two impurities of ${\lambda _1}$ and ${\lambda _2}$ and substitute the Moseley’s law of equation derivatives in the given two impurities by which the atomic number of first and second impurity can be calculated.

Formula used:
Moseley’s law of equation is given by,
$\sqrt \upsilon = a\left( {z - b} \right)$
Where, $\upsilon $ is the frequency of the characteristic X-rays, $a$ and $b$ are constants and $z$ is the atomic number.

Complete step by step solution:
Given, the wavelength of ${K_\alpha }$ lines is ${\lambda _2}$ for target $A$, impurities of ${\lambda _1}$ and ${\lambda _2}$,
$\dfrac{{{\lambda _z}}}{{{\lambda _1}}} = 4$ and $\dfrac{{{\lambda _z}}}{{{\lambda _2}}} = \dfrac{1}{4}$
According to the Moseley’s law of equation,
$\sqrt \upsilon = a\left( {z - b} \right)$
Since, $\upsilon = \dfrac{c}{\lambda }$, substitute this value in above equation, we get
$\sqrt {\dfrac{c}{\lambda }} = a\left( {z - b} \right)$
Since, $a$, $b$ and $c$ are constants. Then,
$\sqrt {\dfrac{1}{\lambda }} = \left( {z - 1} \right)$
Squaring on both sides, we get
$\dfrac{1}{{{\lambda _z}}} = {\left( {z - 1} \right)^2}$, ${\lambda _z}$ denotes the wavelength for particle having atomic number $z$.
So,
$\dfrac{1}{{{\lambda _1}}} = {\left( {{z_1} - 1} \right)^2}$ and $\dfrac{1}{{{\lambda _2}}} = {\left( {{z_2} - 1} \right)^2}$

For First impurity,
$\dfrac{{{\lambda _z}}}{{{\lambda _1}}} = 4$
By substituting the values of ${\lambda _z}$ and ${\lambda _1}$ in above equation, we get
$
  \dfrac{{{{\left( {{z_1} - 1} \right)}^2}}}{{{{\left( {z - 1} \right)}^2}}} = 4 \\
  {\left( {{z_1} - 1} \right)^2} = 4{\left( {z - 1} \right)^2} \\
 $
Taking square root on both sides, we get
$
  \left( {{z_1} - 1} \right) = 2\left( {z - 1} \right) \\
  \left( {{z_1} - 1} \right) = 2z - 2 \\
 $
By performing arithmetic operations, we get
$
  {z_1} = 2z - 2 + 1 \\
  {z_1} = 2z - 1 \\
 $
Thus, the atomic number of the first impurity is $2z - 1$.

For Second impurity,
$\dfrac{{{\lambda _z}}}{{{\lambda _2}}} = \dfrac{1}{4}$
By substituting the values of ${\lambda _z}$ and ${\lambda _2}$ in above equation, we get
$
  \dfrac{{{{\left( {{z_2} - 1} \right)}^2}}}{{{{\left( {z - 1} \right)}^2}}} = \dfrac{1}{4} \\
  {\left( {{z_2} - 1} \right)^2} = \dfrac{{{{\left( {z - 1} \right)}^2}}}{4} \\
 $
Taking square root on both sides, we get
$\left( {{z_2} - 1} \right) = \dfrac{1}{2}\left( {z - 1} \right)$
By performing arithmetic operations, we get
$
  {z_2} = \dfrac{z}{2} - \dfrac{1}{2} + 1 \\
  {z_2} = \dfrac{z}{2} + \dfrac{1}{2} \\
 $
Thus, the atomic number of the second impurity is $\dfrac{{z + 1}}{2}$.

$\therefore$ The atomic number of first impurity is $2z - 1$. The atomic number of second impurity is $\dfrac{{z + 1}}{2}$. Hence, the options (A) and (C) are correct.

Note:
In X-rays tubes, X-rays are generated by accelerating the electrons, creating a vacuum, and bombarding the electrons into a metal target. Except for the tube used in mammography, the target material is usually tungsten.