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.)

Answer 1

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"`