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PhyEasyAdv 2014 • Paper 2

Ratio of Characteristic K Alpha X Ray Wavelengths

Let λCu\lambda_{\text{Cu}} denote the wavelength of the KαK_\alpha X-ray emission line of copper (atomic number 2929) and λMo\lambda_{\text{Mo}} denote the wavelength of the KαK_\alpha X-ray line of molybdenum (atomic number 4242). The ratio λCu/λMo\lambda_{\text{Cu}}/\lambda_{\text{Mo}} is approximately

Options

A

1.991.99

B

2.142.14

Correct
C

0.500.50

D

0.480.48

Topics & Concepts

AtomsX-RaysBohr Model

Step-by-Step Solution

According to Moseley's Law, the frequency ν\nu of the characteristic KαK_\alpha X-ray transition is given by: ν=a(Z−b)\sqrt{\nu} = a(Z - b)

where aa is a proportionality constant, ZZ is the atomic number of the target element, and bb is the screening constant (for the KK-series, b≈1b \approx 1).

Since the wavelength λ\lambda is related to frequency by λ=cν\lambda = \frac{c}{\nu}, we have: 1λ∝(Z−1)2  ⟹  λ∝1(Z−1)2\frac{1}{\lambda} \propto (Z - 1)^2 \implies \lambda \propto \frac{1}{(Z - 1)^2}

Given:

  • For Copper (Cu\text{Cu}): ZCu=29  ⟹  ZCu−1=28Z_{\text{Cu}} = 29 \implies Z_{\text{Cu}} - 1 = 28
  • For Molybdenum (Mo\text{Mo}): ZMo=42  ⟹  ZMo−1=41Z_{\text{Mo}} = 42 \implies Z_{\text{Mo}} - 1 = 41

Therefore, the ratio of their wavelengths is: λCuλMo=(ZMo−1ZCu−1)2=(4128)2\frac{\lambda_{\text{Cu}}}{\lambda_{\text{Mo}}} = \left(\frac{Z_{\text{Mo}} - 1}{Z_{\text{Cu}} - 1}\right)^2 = \left(\frac{41}{28}\right)^2

Calculating the numerical value: λCuλMo=1681784≈2.144≈2.14\frac{\lambda_{\text{Cu}}}{\lambda_{\text{Mo}}} = \frac{1681}{784} \approx 2.144 \approx 2.14

Thus, the correct option is (B).

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