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X-Rays - JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. Get detailed Class 11th & 12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc. eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.     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) $\sqrt{\frac{\mathrm{C}}{\lambda}}=\mathrm{a}(\mathrm{Z}-\mathrm{b})$ $\mathrm{b}=1$ $\sqrt{\frac{\lambda_{\mathrm{Cu}}}{\lambda_{\mathrm{M}_{0}}}}=\left(\frac{Z_{\mathrm{M}_{0}}-1}{Z_{\mathrm{Cu}}-1}\right)$ $\frac{\lambda_{\mathrm{Cu}}}{\lambda_{\mathrm{M}_{0}}}=\left(\frac{41}{28}\right)^{2}=2.14$

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