JEE Advanced 2014 · Paper 2 · Physics Q8

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PhysicsSingle correct+3 / -1Question 8 of 60

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

1.99

2.14

0.50

0.48

Answer and worked solution

Official answer: B

Moseley's law for $K_\alpha$ : $\dfrac1\lambda\propto(Z-1)^2$ .
$\frac{\lambda_{Cu}}{\lambda_{Mo}}=\left(\frac{42-1}{29-1}\right)^2=\left(\frac{41}{28}\right)^2\approx2.14.$

(B)

Topics:Moseley's law X-rays

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JEE Advanced 2014 · Paper 2, Physics question 8