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:
Wavelength of K_β X-ray of argon, λ = 0.36 nm
Energy needed to ionise an argon atom = 16 eV

Energy of K_β X-ray of argon $(E)$ is given by
$E = \frac{1242}{0.36} = 3450 eV$

Energy needed to knock out an electron from K shell
E_K = (3450 + 16) eV
E_K = 3466 eV $\approx$ 3.47 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: Wavelength of Kβ X-ray of argon, λ = 0.36 nm Energy needed to ionise an argon atom = 16 eV Energy of Kβ X-ray of argon E is given by E=12420.36=3450 eV Energy needed to knock out an electron from K shell EK = (3450 + 16) eV EK = 3466 eV ≈ 3.47 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:
Wavelength of K_β X-ray of argon, λ = 0.36 nm
Energy needed to ionise an argon atom = 16 eV

Energy of K_β X-ray of argon $(E)$ is given by
$E = \frac{1242}{0.36} = 3450 eV$

Energy needed to knock out an electron from K shell
E_K = (3450 + 16) eV
E_K = 3466 eV $\approx$ 3.47 keV