Question

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?

Answer 1

Explanation:

It is given that:

Tungsten’s wavelength of X-ray, λ = 21.3 pm.

The required energy to take out electron from the tungsten atom’s L shell, $E_{L}=11.3 \mathrm{keV}$.

The necessary voltage for taking out the electron from the tungsten atom’s L shell, VL = 11.3 kV .

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

$\mathrm{EK}-\mathrm{EL}=\mathrm{hc} \lambda$

where, h = Constant of Planck  

c = light’s speed  

$-\mathrm{EL}+\mathrm{EK}=1242 \mathrm{eV}-10-12 \times \mathrm{nm} 21.3$

$-\mathrm{EL}+\mathrm{EK}=1242 \times 10-9 \mathrm{eV} 21.3 \times 10-12$

$\mathrm{EK}-\mathrm{EL}=58.309 \mathrm{keV}$

$\mathrm{EL}=11.3 \mathrm{keV}$

Therefore,$\mathrm{EK}=69.609 \mathrm{keV}$.

So, the voltage with acceleration across an X-ray tube allowing the Kα X-ray production is shown as $V_{k}=69.609 \mathrm{kV}$.