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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(Z − 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.)
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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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