The $K_{β}$ X -ray of argon has a wavelength of 0.36 nm. The minimum energy needed to ionize an argon atom is 16 eV. Find the energy needed to knock out an electron from the K shell of an argon atom.

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Solution

Given $λ$ = 0.36 nm

E = $\frac{1242}{0.36}$ = 3450 eV

Energy needed to ionize an argon atom

= 16 eV

Energy needed to knock out an electron from K - shell

= (3450+16)eV

= 3466 eV = 3.466 KeV

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The Kβ X -ray of argon has a wavelength of 0.36 nm. The minimum energy needed to ionize an argon atom is 16 eV. Find the energy needed to knock out an electron from the K shell of an argon atom.

Given λ = 0.36 nm E = 12420.36 = 3450 eV Energy needed to ionize an argon atom = 16 eV Energy needed to knock out an electron from K - shell = (3450+16)eV = 3466 eV = 3.466 KeV

The $K_{β}$ X -ray of argon has a wavelength of 0.36 nm. The minimum energy needed to ionize an argon atom is 16 eV. Find the energy needed to knock out an electron from the K shell of an argon atom.

Given $λ$ = 0.36 nm

E = $\frac{1242}{0.36}$ = 3450 eV

Energy needed to ionize an argon atom

= 16 eV

Energy needed to knock out an electron from K - shell

= (3450+16)eV

= 3466 eV = 3.466 KeV