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# The Kα X-rays of Aluminium (Z = 13) and Zinc (Z = 30) Have Wavelengths 887 Pm and 146 Pm Respectively. Use Moseley'S Law √V = A(Z − B) - Physics

ConceptElectromagnetic Spectrum

#### Question

The Kα X-rays of aluminium (Z = 13) and zinc (Z = 30) have wavelengths 887 pm and 146 pm respectively. Use Moseley's  law √v = a(− b) to find the wavelengths of the Kα X-ray of iron (Z = 26).

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

#### Solution

Given:
Wavelength of Kα X-rays of aluminium, λ1 = 887 pm
Frequency of X-rays of aluminium is given by nu_a = c/lambda

nu_a = (3 xx 10^8)/(887 xx 10^-12)

nu_a = 3.382 xx 10^17

nu_a = 33.82 xx 10^16  "Hz"

Wavelength of Kα X-rays of zinc,  lambda_2= 146 pm
Frequency of X-rays of zinc is given by

nu_z = (3 xx 10^8)/(146 xx 10^-12)

nu_z = 0.02055 xx 10^20

nu_z = 2.055 xx 10^18  "Hz"

We know

sqrt(nu) = a(Z-b)

For aluminium,

5.815 xx 10^8 = a(13 - b)  ...(1)

For zinc,

1.4331 xx 10^9 = a(30-b)  ...(2)

Dividing (1) by (2)

(13 - b)/(30 - b) = (5.815 xx 10^-1)/1.4331

= 0.4057

⇒ 30 xx 0.4057 - 0.4057  b = 13 - b

⇒ 12.171 - 0.4057  b + b = 13

b = 0.829/0.5943 = 1.39491

a = (5.815 xx 10^8)/11.33

= 0.51323 xx 10^8 = 5 xx 10^7

For Fe,
Frequency  (nu^') is given by

nu^' = 5× 107 (26 − 1.39)
= 5 × 24.61 × 107
= 123.05 × 107

nu^' = c/lambda

Here, c =  speed of light
lambda = Wavelength of light

therefore  c/lambda = 5141.3 xx 10^14

⇒ lambda = (3 xx 10^8)/(5141.3 xx 10^14)

= 0.000198 xx 10^-5  "m"

= 198 xx 10^-12 = 198  "pm"

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Solution The Kα X-rays of Aluminium (Z = 13) and Zinc (Z = 30) Have Wavelengths 887 Pm and 146 Pm Respectively. Use Moseley'S Law √V = A(Z − B) Concept: Electromagnetic Spectrum.
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