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**Click Here for****JEE main Previous Year Topic Wise Questions of Physics with Solutions****Download eSaral app for free study material and video tutorials.**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]**Ans. (B)
$\sqrt{\frac{\mathrm{C}}{\lambda}}=\mathrm{a}(\mathrm{Z}-\mathrm{b})$
$\mathrm{b}=1$
$\sqrt{\frac{\lambda_{\mathrm{Cu}}}{\lambda_{\mathrm{M}_{0}}}}=\left(\frac{Z_{\mathrm{M}_{0}}-1}{Z_{\mathrm{Cu}}-1}\right)$
$\frac{\lambda_{\mathrm{Cu}}}{\lambda_{\mathrm{M}_{0}}}=\left(\frac{41}{28}\right)^{2}=2.14$