Identify the element that, acting as a target in an x-ray tube, generates Lα radiation of wavelength, λ, equal to 3.07 nm (3.07 x 10-9).
Cu
Ni
Li
Ti
K
Mo
W
To identify the element that generates Lα radiation of a specific wavelength, we can refer to the Moseley's law or the Moseley's equation. Moseley's law states that the square of the wavelength of the emitted X-ray radiation is inversely proportional to the atomic number of the target element.
The equation can be written as:
λ^2 = A(Z - B)
Where λ is the wavelength of the X-ray radiation, Z is the atomic number of the element, and A and B are constants specific to the X-ray tube setup.
In this case, the wavelength is given as 3.07 nm (3.07 x 10^(-9) m). By squaring this value, we get λ^2 = 9.4249 x 10^(-18) m^2.
Now, we need to find the element whose atomic number Z satisfies this equation. We can rearrange the equation to solve for Z:
Z = (λ^2 + B) / A
Since the constants A and B are specific to the X-ray tube setup, we would need to consult a reference or look up the values for the particular X-ray tube being used. However, we can still make a general comparison based on the given options.
Comparing the atomic numbers of the given elements, we find:
Cu: Z = 29
Ni: Z = 28
Li: Z = 3
Ti: Z = 22
K: Z = 19
Mo: Z = 42
W: Z = 74
Out of these options, the element with the atomic number closest to the calculated value would generate the X-ray radiation with the given wavelength. In this case, looking at the options, the element with the closest atomic number is copper (Cu) with an atomic number of 29.
So, based on this comparison, the element that is most likely to generate Lα radiation with a wavelength of 3.07 nm is copper (Cu).