#### 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 × 10^{8} m/s.)

#### Solution

Given:-

Potential of the X-ray tube, V = 40 kV = 40 × 10^{3} V

Energy = 40 × 10^{3} 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"`