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Question: An element with atomic number Z = 11 emits \({{K}_{\alpha }}\)- X-ray of wavelength \(\lambda \). Th...

An element with atomic number Z = 11 emits Kα{{K}_{\alpha }}- X-ray of wavelength λ\lambda . The atomic number which emits Kα{{K}_{\alpha }}- X-ray of wavelength 4λ4\lambda is
A. 4
B. 6
C. 11
D. 44

Explanation

Solution

When an electron t. Use Mosley’s law that gives us the relation between frequency (ν\nu ) of the light emitted and the atomic number of the atom (Z) i.e. ν=a(Z−b)\sqrt{\nu }=a(Z-b)Further use the formula v=cλv=\dfrac{c}{\lambda } to get the relation between Z and λ\lambda .

Formula used:
ν=a(Z−b)\sqrt{\nu }=a(Z-b)
v=cλv=\dfrac{c}{\lambda }

Complete step-by-step answer:
When electrons having high velocity penetrate the surface atoms of the target material, they knock out the tightly bound electrons even from the innermost shells of the atom. When an electron is removed from the atom, a vacancy is created at that place. Electrons from higher shells jump to fill the created vacancies. When an electron jumps from a higher energy orbit to lower energy orbit, it gives out energy in the form of radiation that is equal to the difference in the energy levels. This energy is radiated in the form of X-rays of very small wavelength depending upon the target material.
When the electron penetrates the target, ejects an electron from the K-shell of the target’s atom, a vacancy is created in the K-shell. Immediately, an electron from one of the outer shells, for example L-shell jumps to the K-shell. And in this process it emits an X-ray of energy equal to the energy difference between the two shells. Similarly, if an electron from the M-shell jumps to the K-shell, an X-ray radiation of higher energy is emitted as the energy gap is longer. These radiations are in the form of photons. One electron emits one photon of specific energy. When an electron jumps from L, M, N shells to the K-shell and the emitted X-rays are called , , lines of the K-series of the spectrum.
According to Mosley’s law the frequency of the emitted line depends on the atomic number as
ν=a(Z−b)\sqrt{\nu }=a(Z-b)……..(i),
where ν\nu is frequency of the line and Z is the atomic number.
Here, a is a proportional constant and b is the screening constant. The value of the b depends on interaction of electrons from inner shells. For K-series b=1.
We can write equation (i) as ν=a2(Z−b)2\nu ={{a}^{2}}{{(Z-b)}^{2}}
Therefore, the frequency of Kα{{K}_{\alpha }}- X-ray is ν=a2(Z−1)2\nu ={{a}^{2}}{{(Z-1)}^{2}}
We know that v=cλv=\dfrac{c}{\lambda }, where the speed of light.
Hence, ν=a2(Z−1)2=cλ\nu ={{a}^{2}}{{(Z-1)}^{2}}=\dfrac{c}{\lambda }.
It is given that the wavelength of the X-ray for Z=11 is λ\lambda .
Hence, cλ=a2(11−1)2\dfrac{c}{\lambda }={{a}^{2}}{{(11-1)}^{2}}…….(ii).
It is also said that an atom of atomic number Z emits an X-ray of wavelength 4λ4\lambda .
Hence, c4λ=a2(Z−1)2\dfrac{c}{4\lambda }={{a}^{2}}{{(Z-1)}^{2}}……(iii).
On diving (ii) by (iii) we get,
cλc4λ=a2(11−1)2a2(Z−1)2\dfrac{\dfrac{c}{\lambda }}{\dfrac{c}{4\lambda }}=\dfrac{{{a}^{2}}{{(11-1)}^{2}}}{{{a}^{2}}{{(Z-1)}^{2}}}
⇒(11−1)2(Z−1)2=4\Rightarrow \dfrac{{{(11-1)}^{2}}}{{{(Z-1)}^{2}}}=4
The Tail square roots both the sides.
⇒(10)(Z−1)=±2\Rightarrow \dfrac{(10)}{(Z-1)}=\pm 2
⇒(Z−1)=±5\Rightarrow (Z-1)=\pm 5
This gives us that Z=6 or Z= -4
However, Z is always greater than zero. Hence, Z= -4 is discarded .
Therefore, Z=6.
Hence, the correct option is B.

Note: K-shell consists of 2 electrons. Therefore, if one electron ejected out of the atom, there is still one electron remaining in the K-shell. Due to this an electron in the L-shell will be attracted by the +Ze charge of the nucleus and repelled by –e charge of the electron in the K-shell. Hence the electron in the L-shell will feel a net effective charge of (Z-1)e.