Q4 / 104 · NTA Abhyas (JEE Main) 2020 · Physics · Atomic Physics

The $K_{α}$ line obtained for molybdenum $(Z = 42)$ is $0.71 \overset{\circ}{A}$ . Then, the wavelength of the $K_{α}$ line of copper $(Z = 29)$ is

A
$2.14 \overset{\circ}{A}$
Correct answer
B
$1.52 \overset{\circ}{A}$
C
$1.04 \overset{\circ}{A}$
D
$0.71 \overset{\circ}{A}$

Solution

λ_{min} \propto \frac{1}{{(Z - 1)}^{2}}

∴ \frac{λ_{Cu}}{λ_{Mo}} = \frac{{(Z_{Mo} - 1)}^{2}}{{(Z_{Cu} - 1)}^{2}} = \frac{4 1^{2}}{{(28)}^{2}} = 2.14

∴ λ_{Cu} = 2.14 \times 0.71 \overset{\circ}{A} = 1.52 \overset{\circ}{A}

Hence, the correct answer is option: 'B'.

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