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Kalpha wavelength emitted by an atom of ...

`K_alpha` wavelength emitted by an atom of atomic number `Z=11 " is " lambda`. Find the atomic number for an atom that emits `K_alpha` radiation with wavelength `4lambda`.

A

`Z=6`

B

`Z=4`

C

`Z=11`

D

`Z=44`

Text Solution

Verified by Experts

The correct Answer is:
A
According Moseley.s law, we have `sqrtf = a(Z-b)impliesf=a^2 (Z-b)^2impliesc/lambda =a^2(Z-b)^2` --------(i)
For `K_alpha " line, " b=1`
From Eq. (i) we get
`lambda_2/lambda_1 = ((Z_1-1)^2)/((Z_2-1)^2)implies (4lambda)/lambda = ((11-1)^2)/((Z_2-1)^2)implies Z_2-1 =10/2 implies Z_2 =6`
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