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प्रश्न
The Kα and Kβ X-rays of molybdenum have wavelengths 0.71 A and 0.63 A respectively. Find the wavelength of Lα X-ray of molybdenum.
(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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उत्तर
Given:-
Wavelength of Kα X-rays of molybdenum,`lambda_a = 0.71 "A"`
Wavelength of Kβ X-rays of molybdenum,`lambda_b = 0.63 "A"`
Energy of KαX-rays (`K_a`)is given by
`K_a = E_K - E_L` .....(1)
Energy of Kβ X-rays (`K_β`) is given by
`K_β = E_K - E_L` .....(2)
Energy of La X-ray (`L_a`) is given by
`K_L = E_L - E_M`
Subtracting (2) from (1),
`K_α - K_β = E_M - E_L = -K_L`
or `K_L = K_β - K_α`
`K_L = (3 xx 10^8)/(0.63 xx 10^-10) - (3 xx 10^8)/(0.71 xx 10^-10)`
`K_L = 4.761 xx 10^-18 - 4.225 xx 10^18`
`K_L = 0.536 xx 10^18 "Hz"`
Again , `lambda = (3 xx 10^8)/(0.536 xx 10^-18)`
⇒ `lambda = 5.6 xx 10^-10 = 5.6 Å`
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