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  • An X-ray tube with a copper target is found to emit lines other than those due to copper. The<img alt="\mathrm{\mathrm{K}_\alpha}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathr

An X-ray tube with a copper target is found to emit lines other than those due to copper. The\mathrm{\mathrm{K}_\alpha} line of copper is known to have a wavelength 1.5405 A0 and the other two \mathrm{\mathrm{K}_\alpha} lines observed have wavelengths \mathrm{0.7092 \mathrm{~A}^0} and \mathrm{1.6578 \mathrm{~A}^0} . Identify the impurities (find the value of Z, atomic number).

Option: 1

35 & 24


Option: 2

35 & 28


Option: 3

42 & 19


Option: 4

42 & 28


Answers (1)

best_answer

According to Moseley's equation for \mathrm{\mathrm{K}_a} radiation

\mathrm{\frac{1}{\lambda}=R(Z-1)^2\left[\frac{1}{1^2}-\frac{1}{2^2}\right] \text { where } \lambda=} wavelength of copper

Let \mathrm{\lambda_1 \text { and } \lambda_2} be the two other unknown wavelengths, then

\mathrm{\frac{\lambda_1}{\lambda}=\frac{(Z-1)^2}{\left(Z_1-1\right)^2}=\frac{0.7092}{1.5405}}

\mathrm{\text { Solving we get } Z_1=42 \text {. }}

\mathrm{\text { Similarly } \frac{\lambda_2}{\lambda}=\frac{(28)^2}{\left(Z_2-1\right)^2}=\frac{1.6578}{1.5405}}

\mathrm{\text { Solving we get } Z_2=28 \text {. }}

Posted by

Deependra Verma

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