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**Hint:**The given ${K_\alpha },{K_\beta },{K_\gamma }$ lines are $x - rays$ and hence they behave as electromagnetic waves or EM waves. In this question, we will calculate the energy of the $K$ absorption edge and ${K_\alpha },{K_\beta },{K_\gamma }$ lines using Planck’s formula. Then subtract the energies of ${K_\alpha },{K_\beta },{K_\gamma }$ lines from the energy of $K$ absorption edge to get the energies of the $K,L,M$ orbits in which the $X - rays$ ${K_\alpha },{K_\beta },{K_\gamma }$ lie respectively.

**Formulas Used:**

Energy in an orbit, \[E = \dfrac{{hc}}{\lambda }\]

Where $h$ is Planck’s constant, $c$ is speed of light, $\lambda $ is wavelength

Here, $h$ and $c$ are constant and their product is equal to $1242eV\,nm$ .

**Complete step by step solution:**

In questions like these, convert the wavelength into $nm$ as it is the SI unit for wavelength of $X - rays$. So, wavelength for $K$ absorption edge is at $0.0172nm$, and wavelength of ${K_\alpha },{K_\beta },{K_\gamma }$ lines of $K$ series are \[0.0210nm\], $0.0192nm$ and $0.0108nm$ respectively.

Now, using Planck's formula, we will calculate the energies of each of these.

Planck’s formula says Energy in an orbit, \[E = \dfrac{{hc}}{\lambda }\]

Where $h$ is Planck’s constant, $c$ is speed of light, $\lambda $ is wavelength

Here, $h$ and $c$ are constant and their product is equal to $1242eV\,nm$ .

Therefore, Energy for $K$ absorption edge is \[{E_k} = \dfrac{{hc}}{\lambda }\]

On putting the values, \[{E_k} = \dfrac{{1242}}{{0.0172}} = 72.21KeV\]

Similarly, for ${K_\alpha }$, ${E_{{K_\alpha }}} = \dfrac{{1242}}{{0.0210}} = 59.14KeV$

For ${K_\beta }$, ${E_{{K_\beta }}} = \dfrac{{1242}}{{0.0192}} = 64.69KeV$

For ${K_\gamma }$, ${E_{{K_\gamma }}} = \dfrac{{1242}}{{0.0180}} = 69KeV$

Now we subtract these values from \[{E_k}\] to get the values of energies in the respective orbits.

Energy in $K$ orbit, ${E_K} = {E_{{K_\alpha }}} - {E_k}$

i.e. ${E_K} = 59.14 - 72.21 = - 13.04KeV$

This is option A. Therefore, option A is correct.

Energy in $L$ orbit, ${E_L} = {E_{{K_\beta }}} - {E_k}$

i.e. ${E_L} = 64.69 - 72.21 = - 7.52KeV$

This is option B. Therefore, option B is also correct.

Energy in $M$ orbit, ${E_M} = {E_{{K_\gamma }}} - {E_k}$

i.e. ${E_M} = 69 - 72.21 = - 3.21KeV$

This is option C. Therefore, option C is also correct.

**This means option (A), (B) and (C) are correct.**

**Note:**Convert the units of wavelength in these types of questions otherwise you may end up getting the wrong answer. $h$ and $c$ are constant. $h$ is Planck’s constant, $c$ is speed of light. They have values that are constant throughout the universe. Their product is equal to $1242eV\,nm$. Memorise this value as this may come in handy in questions like these. This will save time.

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