An argon atom has a K_β X-ray wavelength of 0.36 nm . To ionize an...

The $K_\beta$ X-ray is emitted when an electron transitions from the M shell to the K shell. The energy of the emitted $K_\beta$ photon is given by:

$E_{K_\beta} = E_M - E_K$

where $E_M$ and $E_K$ are the energies of the M and K shells, respectively.

The energy of the $K_\beta$ photon can be calculated using the given wavelength $\lambda = 0.36 \text{ nm}$:

$E_{K_\beta} = \dfrac{hc}{\lambda} = \dfrac{1242 \text{ eV}\cdot\text{nm}}{0.36 \text{ nm}} = 3450 \text{ eV} = 3.45 \text{ keV}$

For an argon atom, the outermost electrons are in the M shell ($n=3$). The ionization energy is the energy required to remove an electron from this outermost shell to infinity. Therefore, the binding energy of the M shell is $16 \text{ eV}$, which means:

$E_M = -16 \text{ eV} = -0.016 \text{ keV}$

The energy required to knock out an electron from the K shell is the magnitude of the binding energy of the K shell, which is $-E_K$. Rearranging the transition energy equation:

$-E_K = E_{K_\beta} - E_M$

Substituting the values:

$-E_K = 3.45 \text{ keV} - (-0.016 \text{ keV}) = 3.45 \text{ keV} + 0.016 \text{ keV} = 3.466 \text{ keV}$

Rounding to two decimal places, the required energy is $3.47 \text{ keV}$.