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The wavelength of a certain line in the x-ray spectrum for tungsten \((Z=74\) ) is \(200 \AA\). What would be the wavelength of the same line for platinum \((Z=78)\) ? The screening constant \(a\) is unity. (A) \(179.76 \AA\) (B) \(189.76 \AA\) (C) \(289.76 \AA\) (D) \(379.76 \AA\)

Short Answer

Expert verified
The wavelength of the same line for platinum is \(179.76 \AA\), which is Option (A).

Step by step solution

01

Find the constant b for tungsten

Using Moseley's Law formula, we can calculate the constant b: \(\lambda = \dfrac{b}{(Z - a)^2}\) We know that for tungsten: - \(Z = 74\) - \(\lambda = 200 \AA\) - The screening constant, \(a = 1\) Applying to the formula, \(200 = \dfrac{b}{(74 - 1)^2}\) Solve for \(b\), \(b = 200 \times (73)^2\) Hence, the constant \(b\) is calculated for tungsten.
02

Calculate the wavelength of the same line for platinum

With the constant b found in Step 1, we can now find the wavelength for the same line in the x-ray spectrum of platinum. We know that for platinum: - \(Z = 78\) - The screening constant, \(a = 1\) Using Moseley's Law formula: \(\lambda = \dfrac{b}{(Z - a)^2}\) Substitute the constant \(b\) and values for platinum, \(\lambda = \dfrac{200 \times (73)^2}{(78 - 1)^2}\) After evaluating the expression, we get: \(\lambda = 179.76 \AA\) Hence, the wavelength of the same line for platinum is \(179.76 \AA\), which is Option (A).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

X-ray Spectrum
X-rays are a form of electromagnetic radiation with extremely short wavelengths and high energy. They're found between ultraviolet light and gamma rays on the electromagnetic spectrum. The X-ray spectrum of an element refers to the specific wavelengths emitted when its atoms are excited and emit X-rays. This happens typically when high-energy electrons collide with the target material like tungsten or platinum in X-ray tubes.

In general, the mechanism of X-ray production involves transitions between inner-shell electrons to fill vacancies. These energy transitions produce characteristic X-rays, which are unique to different elements. This unique X-ray spectrum for each element helps in various applications, such as medical imaging and materials analysis. Understanding these unique patterns allows scientists and engineers to identify elements and determine their properties through spectroscopic analysis.

For instance, in the case of tungsten and platinum, the X-ray spectrum is analyzed using characteristics like the wavelength of emitted X-rays. By studying these X-rays, scientists can infer important information about the element's atomic number and its electronic structure.
Atomic Number
The atomic number, denoted by the symbol \(Z\), is one of the most fundamental characteristics of an element. It represents the number of protons in the nucleus of an atom of the element. The atomic number determines the identity of an element and provides insight into its position on the periodic table.

The significance of the atomic number extends to explaining the behavior and characteristics of elements. In the context of X-ray production, Moseley's Law links the frequency (or wavelength) of X-rays emitted by an element to its atomic number. The formula established by Moseley shows a direct relationship between the square root of the X-ray frequency and the atomic number:

- Higher atomic numbers generally produce X-rays with shorter wavelengths due to higher energy transitions
- As observed, substituted values for tungsten \(Z=74\) and platinum \(Z=78\) showcase that a change in atomic number alters the measured wavelength of X-rays

This relationship explains why the emitted wavelengths differ between elements like tungsten and platinum, allowing for identification and comparative analysis based on atomic numbers.
Wavelength Calculation
Calculating the wavelength of X-rays emitted by an element involves using Moseley's Law, which relates the emitted X-ray wavelength to the atomic number of the element and a constant dependent on the element's electronic configuration. Moseley's formula is given by:\[\lambda = \dfrac{b}{(Z - a)^2}\]Here, \(Z\) is the atomic number, \(\lambda\) is the wavelength, and \(a\) is the screening constant, which accounts for the inner electrons partially screening the nucleus charge.

In a practical calculation:
  • The constant \(b\) is first determined using the known X-ray wavelength for a specified element's atomic number.
  • This constant is then used to calculate the wavelength for another element by substituting the new atomic number into the formula.
For tungsten (\(Z=74\)), a wavelength of \(200\, \AA\) provided the means to calculate \(b\). Using this constant, substitution into the formula with platinum's atomic number \(78\) give the resultant wavelength of approximately \(179.76\, \AA\).

This calculation not only demonstrates a method to determine X-ray wavelengths but also emphasizes the predictable nature of X-ray emissions in relation to atomic numbers, making it a powerful tool in physics and chemistry for elemental analysis and identification.

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Most popular questions from this chapter

If doubly ionized lithium atom is hydrogen like with atomic number 3 , the wavelength of radiation required to excite the electron in \(\mathrm{Li}^{\mathrm{H}}\) from the first to the third Bohr orbit and the number of different spectral lines observed in the emission spectrum of the above excited system are (A) \(296 \AA, 6\) (B) \(114 \AA, 3\) (C) \(1026 \AA, 6\) (D) \(8208 \AA, 3\)

If the shortest wavelengths of the continuous spectrum coming out of a Coolidge tube is \(0.1 \AA\), then the de Broglie wavelength of the electron reaching the target metal in the Coolidge tube is approximately ( \(h c=12400 \mathrm{eVA}, h=6.63 \times 10^{-34}\) in MKS, mass of electron \(\left.=9.1 \times 10^{-31} \mathrm{~kg}\right)\) (A) \(0.35 \AA\) (B) \(0.035 \AA\) (C) \(35 \AA\) (D) \(1 \AA\)

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