Question

\[{{K}_{\alpha }}\] wavelength of an unknown element is $0.0709 nm$. Identify the element.
A. Co
B. Cu
C. Mn
D. Mo
E. Sr

Answer 1

Hint: Here we are talking about characteristic wavelength emitted by an element, thus, we have to use Moseley law here. Moseley plotted the x-ray frequencies for about 40 of the elements of the periodic table.

Complete step by step answer:
\[{{K}_{\alpha }}\] x- rays follow a straight line when we plot a graph of atomic number versus the square root of frequency. We are given the wavelength, so first of all find out the frequency first.

\[f=\dfrac{c}{\lambda }\\
\Rightarrow f=\dfrac{3\times {{10}^{8}}}{0.709\times {{10}^{-9}}}\\
\Rightarrow f=4.23\times {{10}^{18}}Hz\]
Now we use the Moseley law here,
\[
Z=1+\sqrt{\dfrac{f}{2.48\times {{10}^{5}}}} \\
\Rightarrow Z=1+\sqrt{\dfrac{4.23\times {{10}^{18}}}{2.48\times {{10}^{5}}}} \\
\Rightarrow Z=1+41.3 \\
\therefore Z=42.3 \\
\]
We know the atomic number is a perfect natural number, so Z comes out to be 42. When we see the periodic table, it comes out to be Molybdenum.

Hence, the correct option is D.

Additional information:
Frequency is the number of complete cycles per second for an oscillating body. It is the characteristic of the source and as the wave propagates through the medium it does not change. It is a constant. While other quantities like wavelength and velocity of the wave depends on the medium and changes.

Note: This calculation for $Z=42$ gives a wavelength of $0.0709 nm$ for the molybdenum K-alpha x-ray whereas the measured value is $0.0707 nm$. So, overall, this is within the domain of error analysis and thus justified. We need to keep in mind that all the units are in standard SI units. The wavelength to be taken in metre and velocity is m/s. The frequency is to be taken in Hertz (hz) or per second.