Calculate the ratio of the wavelength of the K[latex]_\alpha[/latex] line for niobium (Nb) to that for gallium (Ga).Take needed data from the periodic table of Appendix G.

THINK The frequency of an x-ray emission is proportional to (Z – 1)², where Z is the atomic number of the target atom. EXPRESS The ratio of the wavelength λ[latex]_{Nb}[/latex] for the K[latex]_\alpha[/latex] line of niobium to the wavelength λ[latex]_{Ga}[/latex] for the K[latex]_\alpha[/latex] line of gallium is given by [latex]\lambda_{ Nb } / \lambda_{ Ga }=\left(Z_{ Ga }-1\right)^2 /\left(Z_{ Nb}-1\right)^2[/latex] where Z[latex]_{Nb}[/latex] is the atomic number of niobium (41) and Z[latex]_{Ga}[/latex] is the atomic number of gallium (31). Thus, [latex] \lambda_{ Nb } / \lambda_{ Ga }=(30)^2 /(40)^2=9 / 16 \approx 0.563 . [/latex] LEARN The frequency of the K[latex]_\alpha[/latex] line is given by Eq. 40-26: [latex]f=\left(2.46 \times 10^{15} Hz \right)(Z-1)^2 .[/latex] [latex] \begin{aligned} f & =\frac{\Delta E}{h}=\frac{(10.2 eV )(Z-1)^2}{\left(4.14 \times 10^{-15} eV \cdot s \right)} \\ & =\left(2.46 \times 10^{15} Hz \right)(Z-1)^2 . \end{aligned} [/latex] (40-26)

Question 40.39: Calculate the ratio of the wavelength of the Kα line for nio......

Instructor Solution Manual for Fundamentals of Physics [1361627]
Fundamentals of Physics Extended – Instructor’s Solutions Manual [1360823]

Calculate the ratio of the wavelength of the K $_\alpha$ line for niobium (Nb) to that for gallium (Ga).Take needed data from the periodic table of Appendix G.

Step-by-Step

The 'Blue Check Mark' means that this solution was answered by an expert.
Learn more on how do we answer questions.

THINK The frequency of an x-ray emission is proportional to (Z – 1)², where Z is the atomic number of the target atom.

EXPRESS The ratio of the wavelength λ $_{Nb}$ for the K $_\alpha$ line of niobium to the wavelength λ $_{Ga}$ for the K $_\alpha$ line of gallium is given by

$\lambda_{ Nb } / \lambda_{ Ga }=\left(Z_{ Ga }-1\right)^2 /\left(Z_{ Nb}-1\right)^2$

where Z $_{Nb}$ is the atomic number of niobium (41) and Z $_{Ga}$ is the atomic number of gallium (31). Thus, $\lambda_{ Nb } / \lambda_{ Ga }=(30)^2 /(40)^2=9 / 16 \approx 0.563 .$

LEARN The frequency of the K $_\alpha$ line is given by Eq. 40-26:

$f=\left(2.46 \times 10^{15} Hz \right)(Z-1)^2 .$

$\begin{aligned} f & =\frac{\Delta E}{h}=\frac{(10.2 eV )(Z-1)^2}{\left(4.14 \times 10^{-15} eV \cdot s \right)} \\ & =\left(2.46 \times 10^{15} Hz \right)(Z-1)^2 . \end{aligned}$ (40-26)