Question #277693

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]

1
Expert's answer
2021-12-10T12:30:19-0500

Moseley law

ν=z1\sqrt{\nu}=z-1

ν=cλ{\nu}=\frac{c}{\lambda}

cλ=z1\sqrt\frac{c}{\lambda}=z-1

For cobalt λ=λco\lambda=\lambda _{co}

impurities

λ=λx\lambda=\lambda_x

λcoλx=zx1zco1\sqrt{\frac{\lambda_{co}}{\lambda_x}}=\frac{z_x-1}{z_{co}-1}

First impurities

λco=1.785A°λx1=2.285A°λx2=1.537A°\lambda_{co}=1.785A°\\\lambda x_1=2.285A°\\\lambda x_2=1.537A°

λcoλx1=zx11zco1\sqrt{\frac{\lambda_{co}}{{\lambda_x}_1}}=\frac{z_{x1}-1}{z_{co}-1}

1.7852.285=zx11271\sqrt{\frac{1.785}{2.285}}=\frac{z_{x1}-1}{27-1}

Both take square

1.7852.285=zx11271\sqrt{\frac{1.785}{2.285}}=\frac{z_{x1}-1}{27-1}

0.8838=zx11260.8838=\frac{z_{x1}-1}{26}

zx1=23+1=24z_{x1}=23+1=24

Z=24

Cromium



λcoλx2=zx21zco1\sqrt{\frac{\lambda_{co}}{\lambda_{x2}}}=\frac{z_{x2}-1}{z_{co}-1}

Put value

1.7851.537=zx21271\sqrt{\frac{1.785}{1.537}}=\frac{z_{x2}-1}{27-1}

1.07766=zx21261.07766=\frac{z_{x2}-1}{26}

zx2=28+1=29z_{x2}=28+1=29

Z=29

copper



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Comments

BIGIRIMANA Elie
13.12.21, 16:50

thank you to the time you've took for this qwestion

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