In a coolidge tube, the tungsten (Z=74) target is bombarded by electrons. What is the minimum value of the accelerating potential to enable emission of characteristic `K_(alpha)` and `K_(beta)` lines of tungsten. ( The K,L & M levels of tungsten have Binding energies of 69.5, 11.3 & 2.30 keV respectively.)

Calculate the wavelength of K_(alpha) line for the target made of tungsten (Z = 74) .

A cobalt target (Z = 27) is bomberded with electron and the wavelength of its characteristic spectrum are measured . A second , fainter characcteristic spectrum is also found because of an impurity in the target. The wevelength of the K_(alpha) lines are 178.9 pm (cobalt) and 143.5 pm (impurity). What is the impurity?

Find the maximum potential difference which may be applied across an X-ray tube with tungsten target without emitting any characteristics K or L X-ray. The energy levels of the tungsten atom with an electron knocked out are as follows. Cell containing vacancy K L M Energy in keV 69.5 11.3 2.3

Determine the energy of the characteristic X-ray (K_beta) emitted from a tungsten (Z = 74) target when an electron drops from the M-shell (n=3) to a vacancy in the K-shell (n=1)

The wavelength for k_(beta) x-ray of certain, material A is 15.53 pm .If takes 10KeV to remove an electron from M - shell of A .What is the minimum accelerating potential that should be applied across the x -ray tube with target material A . So that a k_(beta) x -ray would be produced? Express answer in powers of 10^(4) ? (Take h=1242ev-nm )

When 50 keV electrons are made incident on a target material, the wavelength of K_(alpha) X–ray line was found to be 0.5Å. When the accelerating potential is increased to 100 kV, then the wavelength of K_(alpha) –line from the same target will be

In a Coolidge tube, the potential difference used to accelerate the electrons is increased from 24. 8 kV to 49.6 kV . As a result, the difference between the wavelength of K_(alpha) -line and minimum wavelength becomes two times. The initial wavelength of the K_(alpha) line is [Take (hc)/(e) = 12 .4 kV Å ]