Answer to Question #277693 in Atomic and Nuclear Physics for elie

Question

X-ray from a certain cobalt target tube are composed of the strong 𝐾 series

of Cobalt and weak 𝐾 lines due to impurities. The wavelengths of 𝐾𝛼 are

1.785𝐴

𝑜

for Cobalt and 2.285𝐴

𝑜

and 1.537𝐴

𝑜

for impurities. (a) What

elements are they? [For the 𝐾-series, a=1 in Moseley’s law]

Answer 1

Moseley law

$\sqrt{\nu}=z-1$

${\nu}=\frac{c}{\lambda}$

$\sqrt\frac{c}{\lambda}=z-1$

For cobalt $\lambda=\lambda _{co}$

impurities

$\lambda=\lambda_x$

$\sqrt{\frac{\lambda_{co}}{\lambda_x}}=\frac{z_x-1}{z_{co}-1}$

First impurities

$\lambda_{co}=1.785A°\\\lambda x_1=2.285A°\\\lambda x_2=1.537A°$

$\sqrt{\frac{\lambda_{co}}{{\lambda_x}_1}}=\frac{z_{x1}-1}{z_{co}-1}$

$\sqrt{\frac{1.785}{2.285}}=\frac{z_{x1}-1}{27-1}$

Both take square

$\sqrt{\frac{1.785}{2.285}}=\frac{z_{x1}-1}{27-1}$

$0.8838=\frac{z_{x1}-1}{26}$

$z_{x1}=23+1=24$

Z=24

Cromium

$\sqrt{\frac{\lambda_{co}}{\lambda_{x2}}}=\frac{z_{x2}-1}{z_{co}-1}$

Put value

$\sqrt{\frac{1.785}{1.537}}=\frac{z_{x2}-1}{27-1}$

$1.07766=\frac{z_{x2}-1}{26}$

$z_{x2}=28+1=29$

Z=29

copper