Moseley's Law

The frequency $\nu$ of a characteristic X-ray of an element is related to its atomic number $Z$ by \begin{align} \sqrt{\nu}=a(Z-b), \end{align} where $a$ and $b$ are constants called proportionality and screening (or shielding) constants. For $K$ series, the value of $a$ is $\sqrt{3Rc/4}$ and that of $b$ is 1. Here $R$ is Rydberg's constant and $c$ is speed of light (as in Bohr's model). For $L$ series, the value of $a$ is $\sqrt{5Rc/36}$ and $b$ is 7.4. The relation and values of $a$ and $b$ are experimentally determined by Henry Moseley.

Solved Problems on Moseley's Law

Problem from IIT JEE 2003

Characteristic X-rays of frequency ${4.2\times{10}^{18}}$  Hz are produced when transitions from $L$-shell to $K$-shell take place in a certain target material. Use Moseley's law to determine the atomic number of the target material. (Rydberg's constant $={1.1\times{10}^{7}}\;\mathrm{m^{-1}}$.)

Solution:

The characteristic X-ray is emitted when an electron in $L$ shell makes a transition to the vacant state in $K$ shell. In Moseley's equation, \begin{align} \sqrt{\nu}=a(Z-b), \end{align} the parameter $b\approx 1$ for this transition because electron from $L$ shell finds nuclear charge $Ze$ shielded by remaining one electron in $K$ shell i.e., effective nuclear charge is $(Z-1)e$. Thus, by substituting values, \begin{align} \frac{1}{\lambda}&=\frac{\nu}{c}=\frac{4.2\times{10}^{18}}{3\times{10}^{8}}\nonumber\\ &=R(Z-1)^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right] \nonumber\\ &=1.1\times{10}^{7} (Z-1)^2\left[\frac{1}{1^2}-\frac{1}{2^2}\right],\nonumber \end{align} which gives, $Z=42$. Moseley's law played key role in arrangement of elements in the periodic table and to find many new (missing) elements.

Problem from IIT JEE 2014

If $\lambda_\mathrm{Cu}$ is the wavelength of $K_\alpha$ X-ray line of copper (atomic number 29) and $\lambda_\mathrm{Mo}$ is the wavelength of the $K_\alpha$ X-ray line of molybdenum (atomic number 42), the the ratio $\lambda_\mathrm{Cu}/\lambda_\mathrm{Mo}$ is close to

1. 1.99
2. 2.14
3. 0.50
4. 0.48

Solution:

The wavelength of $K_\alpha$ X-ray line is related to atomic number $Z$ by Moseley's Formula \begin{align} \frac{1}{\lambda}&=R(Z-1)^2\left[\frac{1}{1^2}-\frac{1}{2^2}\right]\nonumber\\ &=\frac{3}{4}R(Z-1)^2. \nonumber \end{align} Substitute the value of $Z$ to get \begin{align} \frac{\lambda_\mathrm{Cu}}{\lambda_\mathrm{Mo}}=\frac{(Z_\mathrm{Mo}-1)^2}{(Z_\mathrm{Cu}-1)^2}=\frac{(41)^2}{(28)^2}=2.14.\nonumber \end{align} The elements with higher atomic number (molybdenum in this example) gives high energy X-rays (short wavelengths).

Problem from IIT JEE 2008

Which of the following statements is wrong in the context of X-rays generated from a X-ray tube?

1. Wavelength of characteristic X-rays decreases when the atomic number of the target increases.
2. Cut-off wavelength of the continuous X-rays depends on the atomic number of the target.
3. Intensity of the characteristic X-rays depends on the electrical power given to the X-ray tube.
4. Cut-off wavelength of the continuous X-rays depends on the energy of the electrons in the X-ray tube.

Solution:

The frequency $\nu$ of characteristic X-rays is related to atomic number $Z$ by Moseley's law, \begin{align} \sqrt{\nu}=a(Z-b),\nonumber \end{align} which gives \begin{align} \lambda=\frac{c}{\nu}=\frac{c}{a^2(Z-b)^2}.\nonumber \end{align} Thus, the wavelength of emitted X-rays decreases with increase in $Z$. The cut-off wavelength of continuous X-rays corresponds to maximum energy of electron in X-ray tube. It is given by \begin{align} hc/\lambda=eV,\nonumber \end{align} where $V$ is the accelerating potential. The intensity of X-rays depends on the number of electrons striking the target per second, which, in turn, depends on the electrical power given to the X-ray tube as energy of each electron is $eV$.

Questions on Moseley's Law

Question: If 178.5 pm is the wavelength of X-ray line of copper (atomic number 29) and 71 pm is the wavelength of the X-ray line of molybdenum (atomic number 42) then the value of a and b in Moseley's equation are

A. $a=7\times 10^7\, \mathrm{Hz}^{1/2}$, b=1.0
B. $a=5\times 10^7\, \mathrm{Hz}^{1/2}$, b=7.4
C. $a=5\times 10^7\, \mathrm{Hz}^{1/2}$, b=1.4
D. $a=7\times 10^7\, \mathrm{Hz}^{1/2}$, b=7.4

Question: Moseley's Law for characteristic X-rays is $\sqrt{\nu}=a(Z-b)$. In this formula

A. both a and b are independent on the material.
B. a is independent but b depends on the material.
C. b is independent but a depends on the material.
D. both a and b depend on the material.

Related Topic

1. Characteristic and Continuous X-rays | Problems | IIT JEE

References and External Links

1. Concepts of Physics Part 2 by HC Verma (Link to Amazon)
2. IIT JEE Physics by Jitender Singh and Shraddhesh Chaturvedi
3. X-ray Fluorescence and Moseley’s Law (pdf article on X-rays, Moseley's Law and Moseley's Experiments)
4. The Physical (in)significance of Moseley's Screening Parameters, (journal article in pdf by K Razi Naqvi)
5. One hundred years of Moseley’s law: An undergraduate experiment with relativistic effects (pdf)