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

# P the Wavelength of Kα X-ray of Tungsten is 21.3 Pm. It Takes 11.3 Kev to Knock Out an Electron from the L Shell of a Tungsten Atom. What Should Be the Minimum Accelerating - Physics

Sum

The wavelength of Kα X-ray of tungsten is 21.3 pm. It takes 11.3 keV to knock out an electron from the L shell of a tungsten atom. What should be the minimum accelerating voltage across an X-ray tube having tungsten target which allows production of Kα X-ray?

(Use Planck constant h = 6.63 × 10-34 Js= 4.14 × 10-15 eVs, speed of light c = 3 × 108 m/s.)

#### Solution

Given:-

Wavelength of X-ray of tungsten,

lambda = 21.3 pm

Energy required to take out electron from the L shell of a tungsten atom, EL = 11.3 keV

Voltage required to take out electron from the L shell of a tungsten atom, VL = 11.3 kV

Let EK and EL be the energies of K and L, respectively.

E_K - E_L = (hc)/lambda

Here,

h = Planck's constant

c = Speed of light

E_K - E_L = (1242  "eV" - "nm")/(21.3 xx 10^-12)

E_K - E_L = (1242 xx 10^-9  "eV")/(21.3 xx 10^-12)

E_K - E_L = 58.309  "keV"

E_L = 11.3  "keV"

therefore E_K  = 69.609  "keV"

Thus, the accelerating voltage across an X-ray tube that allows the production of Kα X-ray is given by

VK = 69.609 kV

Is there an error in this question or solution?

Share