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For a certain metal, the $K$ absorption edge is at $0.172 A^o$. The wavelength of ${K_\alpha },{K_\beta },{K_\gamma }$ lines of $K$ series are \[0.210 A^o \], $0.192 A^o$ and $0.108 A^o$ respectively. The energies of $K,L,M$ orbits are ${E_K},{E_L}$ and ${E_M}$ respectively. Then:
A) ${E_K} = - 13.04\,keV$
B) ${E_L} = - 7.52\,keV$
C) ${E_M} = - 3.21\,keV$
D) ${E_K} = 13.04\,keV$

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Last updated date: 18th Jun 2024
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Answer
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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.