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# A Certain Element Emits Kα X-ray of Energy 3.69 Kev. Use the Data from the Previous Problem to Identify the Element. - Physics

ConceptElectromagnetic Spectrum

#### Question

A certain element emits Kα X-ray of energy 3.69 keV. Use the data from the previous problem to identify the element.

(Use Planck constant h = 6.63 × 10-34 Js= 4.14 × 10-15 eVs, speed of light c = 3 × 108 m/s.)

#### Solution

Given:
Energy of Kα X-ray, E = 3.69 keV = 3690 eV

Wavelength (λ) is given by

λ = (hc)/E

λ = 1242/3690

λ = 0.33658  "nm"

⇒ λ = 0.34 xx 10^-9  "m"

From Moseley's equation,

sqrt(c/lambda) = a(Z - b)

Here, c = speed of light
lambda = wavelength of light
Z = atomic number of element

On substituting the respective values,

sqrt((3 xx 10^8)/(0.34 xx 10^-9)) = 5 xx 10^7(Z - 1.39)

⇒ sqrt(8.82 xx 10^17) = 5 xx 10^7(Z - 1.39)

⇒ 9.39 xx 10^8 = 5 xx 10^7(Z - 1.39)

⇒ 93.9/5 = Z - 1.39

⇒ Z = 93.9/5 + 1.39

= 20.17 = 20

So, the element is calcium.

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Solution A Certain Element Emits Kα X-ray of Energy 3.69 Kev. Use the Data from the Previous Problem to Identify the Element. Concept: Electromagnetic Spectrum.
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