X-Rays – JEE Advanced Previous Year Questions with Solutions
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Q. If $\lambda_{\mathrm{Cu}}$ is the wavelength of $\mathrm{K}_{\alpha}$ X-ray line of copper (atomic number 29) and $\lambda_{\mathrm{Mo}}$ is the wavelength of the $\mathrm{K}_{\alpha}$ X-ray line of molybdenum (atomic number 42), then the ratio is close to :-(A) 1.99                (B) 2.14               (C) 0.50                (D) 0.48 [JEE Advance-2014]

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Sol. (B)$\sqrt{\frac{\mathrm{C}}{\lambda}}=\mathrm{a}(\mathrm{Z}-\mathrm{b})$$\mathrm{b}=1$$\sqrt{\frac{\lambda_{\mathrm{Cu}}}{\lambda_{\mathrm{M}_{0}}}}=\left(\frac{Z_{\mathrm{M}_{0}}-1}{Z_{\mathrm{Cu}}-1}\right)$$\frac{\lambda_{\mathrm{Cu}}}{\lambda_{\mathrm{M}_{0}}}=\left(\frac{41}{28}\right)^{2}=2.14$