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Question

X-rays emitted from a copper target and a molybdenum target are found to contain a line of wavelength 22.85 nm attribute to the Kα lines of an impure element. The Kα lines of copper (Z =29) and molybdenum (Z =42) have a wavelength 15.42 nm and 7.12 nm respectively. Using Moseley's law, v1/2=a(Zb), calculate the atomic number of the impure element:

A
24
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B
28
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C
30
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D
32
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Solution

The correct option is A 24
The moseley's law is ν=cλ=a(Zb)
For copper, 3×10815.42×109m=a(29b)......(1)
For, Mo, 3×1087.12×109m=a(42b)......(2)
Subtract equation (1) from equation (2)
6.58×107=42a29a=13a
a=5.06×106
Substitute the value of a in equation (1)
3×10815.42×109m=5.06×106(29b)
b=1.44
For impurity, 3×10822.852×109m=5.06×106×Z(291.44)
Z=24

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