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

Atoms and Nuclei PYQ — JEE Advanced 2014 · Shift 2 · Q46 with Solution

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Correct answer: B

For a $K_\alpha$ X-ray, the electron transition is from $n=2$ to $n=1$ .

The frequency is approximately

\nu \propto (Z-b)^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right)

where $b\approx 1$ for the $K$ -series screening.

Since

\frac{1}{1^2}-\frac{1}{2^2}=1-\frac14=\frac34

is the same for both elements, we get

\nu \propto (Z-1)^2

And because

\lambda = \frac{c}{\nu},

we have

\lambda \propto \frac{1}{(Z-1)^2}

So,

\frac{\lambda_{\rm Cu}}{\lambda_{\rm Mo}}=\frac{(Z_{\rm Mo}-1)^2}{(Z_{\rm Cu}-1)^2}

For copper, $Z_{\rm Cu}=29$ .

For molybdenum, $Z_{\rm Mo}=42$ .

Thus,

\frac{\lambda_{\rm Cu}}{\lambda_{\rm Mo}}=\frac{(42-1)^2}{(29-1)^2} = \frac{41^2}{28^2}

41^2=1681, \qquad 28^2=784

Hence,

\frac{\lambda_{\rm Cu}}{\lambda_{\rm Mo}}=\frac{1681}{784}\approx 2.14

\frac{\lambda_{\rm Cu}}{\lambda_{\rm Mo}}\approx 2.14

So the correct option is B.

Stored correct answer: B

This matches the derived answer.