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Question
An X-ray tube operates at 40 kV. Suppose the electron converts 70% of its energy into a photon at each collision. Find the lowest there wavelengths emitted from the tube. Neglect the energy imparted to the atom with which the electron collides.
(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:-
Potential of the X-ray tube, V = 40 kV = 40 × 103 V
Energy = 40 × 103 eV
Energy utilised by the electron is given by `E = 70/100 xx 40 xx 10^3 = 28 xx 10^3 "eV"`
Wavelength (`lambda`) is given by
`lambda = (hc)/E`
Here,
h = Planck's constant
c = Speed of light
E = Energy of the electron
`therefore lambda = (hc)/E`
`⇒ lambda = (1242 "eV" - "nm")/(28 xx 10^3 "eV")`
`⇒ lambda = (1242 xx 10^-9 "eV")/(28 xx 10^3 "eV")`
`⇒ lambda = 44.35 xx 10^-12`
`⇒ lambda = 44.35 "pm"`
For the second wavelength,
E = 70% (Leftover energy)
`= 70/100 xx (40-28)10^3`
`= 70/100 xx 12 xx 10^3`
`= 84 xx 10^2 "eV"`
And
`lambda = (hc)/E = 1242/(8.4 xx 10^3)`
`= 147.86 xx 10^-3 "nm"`
`= 147.86 "pm" = 148 "pm"`
For the third wavelength,
`E = 70/100(12 - 8.4) xx 10^3`
`= 7 xx 3.6 xx 10^2 = 25.2 xx 10^2 "eV"`
And ,
`lambda = (hc)/E = 1242/(25.2 xx 10^2)`
`= 49.2857 xx 10^-2`
`= 493 "pm"`
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