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Question

Xrays emitted from a copper target and a molybdenum target are found to contain a line of wavelength 22.85 nm attributed to the Kα line of an impurity element. The Kα lines of copper (Z=29) and molybdenum (Z=42) have wavelengths 15.42 nm and 7.12 nm, respectively. The atomic number of the impurity element is

A
22
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B
23
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C
24
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D
25
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Solution

The correct option is C 24
For Kα line, 1/λ=3R(Z1)2/4

So, λcopper/λimpurity=(Zimpurity1)2(Zcopper1)2 15.42/22.85=(Zimpurity1)2282 (Zimpurity1)2=529 $

Zimpurity=24

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