Question 72: Atoms & Nuclei

Question

If $$\lambda$$Cu is the wavelength of K$$\alpha$$ X-ray line of copper (atomic number 29) and $$\lambda$$Mo is the wavelength of the K$$\alpha$$ X-ray line of molybdenum (atomic number 42), then the ratio $$\lambda$$Cu/$$\lambda$$Mo is close to

Accepted Answer

The wavelength of the K
$$\alpha$$
X-ray line is related to the atomic number of the element by Moseley's law, which can be expressed as:
$$ \sqrt{
u} = A(Z - B) $$
Here, $$
u$$ is the frequency of the emitted X-ray, $$Z$$ is the atomic number, and $$A$$ and $$B$$ are constants. Given that the wavelength $$\lambda$$ is inversely proportional to the frequency $$
u$$ (since $$
u = \frac{c}{\lambda}$$, where $$c$$ is the speed of light), we can rewrite this equation in terms of wavelength:
$$ \lambda \propto \frac{1}{(Z - B)^2} $$
Now, let's consider the ratio of the wavelengths for copper (Cu) and molybdenum (Mo):
$$ \frac{\lambda_{Cu}}{\lambda_{Mo}} = \left(\frac{Z_{Mo} - B}{Z_{Cu} - B}ight)^2 $$
For the K
$$\alpha$$
X-ray line, the constant $$B$$ can be approximated as 1. Using the atomic numbers $$Z_{Cu} = 29$$ and $$Z_{Mo} = 42$$, we can plug these values into the ratio:
$$ \frac{\lambda_{Cu}}{\lambda_{Mo}} = \left(\frac{42 - 1}{29 - 1}ight)^2 = \left(\frac{41}{28}ight)^2 $$
Calculating the above expression:
$$ \frac{41}{28} \approx 1.464 $$
$$ \left(1.464ight)^2 \approx 2.14 $$
Therefore, the ratio $$\frac{\lambda_{Cu}}{\lambda_{Mo}}$$ is close to 2.14. The correct option is:
Option B
2.14

Answer

1.99

Answer

2.14

Answer

0.50

Answer

0.48

Atoms & Nuclei

If $\lambdaCu is the wavelength of K\alpha X-ray line of copper (atomic number 29) and \lambdaMo is the wavelength of the K\alpha X-ray line of molybdenum (atomic number 42), then the ratio \lambdaCu/\lambda$Mo is close to

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