If $\lambda_{Cu}$ is the wavelength of $K_{\alpha}$ X-ray line of copper (atomic number 29) and $\lambda_{Mo}$ is the wavelength of the $K_{\alpha}$ X-ray line of molybdenum (atomic number 42), then the ratio $\lambda_{Cu} / \lambda_{Mo}$ is close to

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Modern Physics

λ_{C u}

is the wavelength of

K_{α}

X-ray line of copper (atomic number 29) and

λ_{M o}

is the wavelength of the

K_{α}

X-ray line of molybdenum (atomic number 42), then the ratio

λ_{C u} / λ_{M o}

is close to

(A)1.99
(B)2.14✓ Correct
(C)0.50
(D)0.48

Answer: (B)

Solution

Using Moseley's law for

K_{α}

X-rays:

ν = a (Z - b)

, where

b = 1

for

K_{α}

transitions.

Since

ν = \frac{c}{λ}

, we have

\frac{1}{λ} \propto (Z - 1)

.

Therefore:

\frac{λ _{C u}}{λ _{M o}} = (\frac{Z _{M o} - 1}{Z _{C u} - 1})^{2} = (\frac{42 - 1}{29 - 1})^{2} = (\frac{41}{28})^{2}

.

Calculating:

(\frac{41}{28})^{2} = (1.464)^{2} \approx 2.14

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Single correct option

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