Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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# 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 - Physics

ConceptElectromagnetic Spectrum

#### 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.)

#### 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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Solution 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 Concept: Electromagnetic Spectrum.
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