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प्रश्न
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.)
उत्तर
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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