If is the wavelength of X-ray line of copper (atomic number 29) and is the wavelength of the X-ray line of molybdenum (atomic number 42), then the ratio is close to [2014]

Question

If $λ_{Cu}$ is the wavelength of $K_{α}$ X-ray line of copper (atomic number 29) and $λ_{Mo}$ is the wavelength of the $K_{α}$ X-ray line of molybdenum (atomic number 42), then the ratio $\frac{λ_{Cu}}{λ_{Mo}}$ is close to [2014]

Smart Achievers · Accepted Answer

(2)

$∵$ $K_{α}$

For K-series, $b = 1$ and for $K_{\propto}$ transition $n_{f} = 2$ to $n_{i} = 1$ ,

$∴$ $\frac{1}{λ} \propto {(z - 1)}^{2}$

$∵$ $\frac{λ_{Cu}}{λ_{Mo}} = \frac{{(Z_{Mo} - 1)}^{2}}{{(Z_{Cu} - 1)}^{2}} = {(\frac{42 - 1}{29 - 1})}^{2} = {(\frac{41}{28})}^{2} = 2.14$

Solution

Q.
If $λ_{Cu}$ is the wavelength of $K_{α}$ X-ray line of copper (atomic number 29) and $λ_{Mo}$ is the wavelength of the $K_{α}$ X-ray line of molybdenum (atomic number 42), then the ratio $\frac{λ_{Cu}}{λ_{Mo}}$ is close to [2014]

1 1.99

2 2.14

3 0.50

4 0.48

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