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Question: An element of atomic number \(9\) units \({{K}_{\alpha }}\) X-ray of wavelength \(\lambda \). Find t...

An element of atomic number 99 units Kα{{K}_{\alpha }} X-ray of wavelength λ\lambda . Find the atomic number of the element which emits Kα{{K}_{\alpha }} X-ray of wavelength 4λ4\lambda .

Explanation

Solution

Hint We use the formula
1λ=R(Z−1)2(1n12−1n22)\dfrac{1}{\lambda }=R{{\left( Z-1 \right)}^{2}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right) …………………(1)
Where λ=\lambda = Wavelength, Z=Z= Atomic number, and R=R= Rydberg constant

Complete step-by-step solution :
K−alpha:−KαK-alpha:- {{K}_{\alpha }} is typically by for the strongest X-Ray spectral lines for an element bombarded with energy sufficient to cause maximally intense X-Ray emission.
Moseley equation:- It states that the frequency of the spectral line in the characteristic X-Ray spectrum is directly proportional to the square of the atomic number of the element considered.
f=a(Z−b)\sqrt{f}=a\left( Z-b \right)
According to the question,
Z1=9 λ1=λ \begin{aligned} & {{Z}_{1}}=9 \\\ & {{\lambda }_{1}}=\lambda \\\ \end{aligned}
Then by the equation (1)
1λ=R(9−1)2(1n12−1n22).............(2) Z2=Z λ2=4λ \begin{aligned} & \dfrac{1}{\lambda }=R{{\left( 9-1 \right)}^{2}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right).............(2) \\\ & {{Z}_{2}}=Z \\\ & {{\lambda }_{2}}=4\lambda \\\ \end{aligned}
14λ=R(Z−1)2(1n12−1n22)...............(3)\dfrac{1}{4\lambda }=R{{\left( Z-1 \right)}^{2}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)...............(3)
Equation (2) is divided by equation (3)
1λ14λ=R(9−1)R(Z−1)(1n12−1n22)(1n12−1n22)\dfrac{\dfrac{1}{\lambda }}{\dfrac{1}{4\lambda }}=\dfrac{R\left( 9-1 \right)}{R\left( Z-1 \right)}\dfrac{\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)}{\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)}
4=(8)2(Z−1)24=\dfrac{{{\left( 8 \right)}^{2}}}{{{\left( Z-1 \right)}^{2}}}
(Z−1)2=8×84 (Z−1)2=16 \begin{aligned} & {{\left( Z-1 \right)}^{2}}=\dfrac{8\times 8}{4} \\\ & {{\left( Z-1 \right)}^{2}}=16 \\\ \end{aligned}
Z−1=4 Z=4+1 Z=5 \begin{aligned} & Z-1=4 \\\ & Z=4+1 \\\ & Z=5 \\\ \end{aligned}

Note: Student think that when the wavelength are different, then the transition are also different but transition takes place for both between same states.