By applying Moseley's law with b = 1 , calculate the frequency of the K_α X-ray of La ( Z = 57 ) given that the frequency of the K_α X-ray of Cu ( Z = 29 ) is 1.88 × 10^ 18 Hz .

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Quantrex Academy · Accepted Answer

Correct answer: C. 7.52 × 10^ 18 Hz

According to Moseley's law, the frequency

\nu

of

K_\alpha

X-ray is given by:

\sqrt{\nu} = a(Z - b)

Given

b = 1

, we have

\nu \propto (Z - 1)^2

.

For Cu (

Z = 29

):

\nu_{\text{Cu}} \propto (29 - 1)^2 = 28^2

For La (

Z = 57

):

\nu_{\text{La}} \propto (57 - 1)^2 = 56^2

Taking the ratio of the frequencies:

\dfrac{\nu_{\text{La}}}{\nu_{\text{Cu}}} = \left(\dfrac{56}{28}\right)^2 = 2^2 = 4

\nu_{\text{La}} = 4 \times \nu_{\text{Cu}}

Substituting the given value

\nu_{\text{Cu}} = 1.88 \times 10^{18} \text{ Hz}

:

\nu_{\text{La}} = 4 \times 1.88 \times 10^{18} \text{ Hz} = 7.52 \times 10^{18} \text{ Hz}

By applying Moseley's law with b = 1 , calculate the frequency of...

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