Solveeit Logo

Question

Question: An element with atomic number Z = 11 emits X-ray of wavelength λ. The atomic number of the element w...

An element with atomic number Z = 11 emits X-ray of wavelength λ. The atomic number of the element which emits Kα X-ray of wavelength 4λ is?
A. 11
B. 44
C. 6
D. 5

Explanation

Solution

As we all know that, the Moseley’s law is an empirical law related to atoms that emit XX rays. According to Moseley’s experiment, the square root of the frequencies of various elements is plotted against the position number in the periodic table.

Complete step by step solution:
We all know that in an X-ray spectrum, the frequency of the spectral line is directly proportional to the square root of the atomic number (Z) of the element under study.
v=a(Zb)\sqrt v = a\left( {Z - b} \right)…… (I)
As we can see that here aa and bb are constants depending upon the subsequent spectral line and vv is the frequency
We can substitute v=cλv = \dfrac{c}{\lambda } in equation (I) and here cc is the speed of light and λ\lambda is the wavelength of the spectral line. Now after substituting we will now get the following result as,
cλ=a(Zb)\Rightarrow \sqrt {\dfrac{c}{\lambda }} = a\left( {Z - b} \right)
cλ=a2(Zb)2\Rightarrow \dfrac{c}{\lambda } = {a^2}{\left( {Z - b} \right)^2}
So, we can write it for two different elements using 1 and 2 as subscripts.
cλ1=a2(Z1b)2\Rightarrow \dfrac{c}{{{\lambda _1}}} = {a^2}{\left( {{Z_1} - b} \right)^2}…… (II)
cλ2=a2(Z2b)2\Rightarrow \dfrac{c}{{{\lambda _2}}} = {a^2}{\left( {{Z_2} - b} \right)^2}…… (III)
Here λ1{\lambda _1} is for the given Z1=11{Z_1} = 11 and λ2{\lambda _2} is for the unknown Z2{Z_2} .
So we can divide equation (I) by equation (II) to get the relationship between
λ1{\lambda _1} and λ2{\lambda _2}.
λ1λ2=(Z21)2(Z11)2\therefore \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{{{{\left( {{Z_2} - 1} \right)}^2}}}{{{{\left( {{Z_1} - 1} \right)}^2}}}…… (IV)
We can here now substitute λ1=λ{\lambda _1} = \lambda , Z1=11{Z_1} = 11, λ2=4λ{\lambda _2} = 4\lambda in equation (IV) to find the value of Z2{Z_2}.
λ4λ=(Z21)2(111)2\Rightarrow \dfrac{\lambda }{{4\lambda }} = \dfrac{{{{\left( {{Z_2} - 1} \right)}^2}}}{{{{\left( {11 - 1} \right)}^2}}}
(Z21)2=25\Rightarrow {\left( {{Z_2} - 1} \right)^2} = 25
Z2=6\therefore {Z_2} = 6

\therefore Hence, Option (C) is correct.

Note:
As we all know that using Moseley’s law, gases like krypton and argon, Nickel and cobalt were placed in a proper way in a periodic table. We also know that many new elements like Technetium, Rhodium, etc. were discovered using this law.