Answer
$\boxed{\text{percentage deviation}=-6.78\%}$
Work Step by Step
The energy associated with an atomic shell of an element having atomic number $Z$ is given by
$E_n=-\frac{me^4}{8\epsilon_0^2h^2n^2}(Z-1)^2$ ......................$(1)$
The $K_{\alpha}$ x-ray photon arises when an electron makes a transition from the $L$ shell (with $n=2$ and energy $E_2$) to the $K$ shell (with $n=1$ and energy $E_1$). Thus, using Eq. $1$, we may write the energy change as
$\Delta E=E_2-E_1$
or, $\Delta E=-\frac{me^4}{8\epsilon_0^2h^22^2}(Z-1)^2-\frac{-me^4}{8\epsilon_0^2h^21^2}(Z-1)^2$
or, $\Delta E=\frac{me^4}{8\epsilon_0^2h^2}\Big[1-\frac{1}{4}\Big](Z-1)^2$
or, $\Delta E=\frac{3me^4}{32\epsilon_0^2h^2}(Z-1)^2$
Then the frequency $f$ of the $K_{\alpha}$ line is
$f=\frac{\Delta E}{h}$
or, $f=\frac{3me^4}{32\epsilon_0^2h^3}(Z-1)^2$
or, $\sqrt f=C(Z-1)$ .....................$(2)$
where, $C=\sqrt {\frac{3me^4}{32\epsilon_0^2h^3}}$ is a constant.
Substituting the the known values in $C=\sqrt {\frac{3me^4}{32\epsilon_0^2h^3}}$, we obtain
$C=\sqrt {\frac{3\times 9.11\times 10^{-31}\times (1.6\times 10^{-19})^4}{32\times(8.85\times 10^{-12})^2\times(6.63\times10^{-34})^3}}\;Hz^{1/2}$
$C=4.9519\times10^{7}\;Hz^{1/2}$
Now, the theoretically calculated energy of $K_{\alpha}$ x-ray photon is given by
$E_{theory}=hf$
or, $E_{theory}=hC^2(Z-1)^2$ (using Eq. $2$) .....................$(3)$
Substituting $Z=7$ for nitrogen (N), we obtain
$E_{theory}=6.63\times10^{-34}\times(4.9519\times10^{7})^2\times(7-1)^2\;J$
$E_{theory}=\frac{6.63\times10^{-34}\times(4.9519\times10^{7})^2\times(7-1)^2}{1.6\times10^{-19}}\;eV$
$E_{theory}=365.796\;eV$
But the experimentally measured energy of $K_{\alpha}$ x-ray photon for N target is $E_{exp}=392.4\;eV$
Therefore, the percentage deviation between $E_{theory}$ and $E_{exp}$ for N target is given by
$\text{percentage deviation}=\frac{E_{theory}-E_{exp}}{E_{exp}}\times100=\frac{365.796-392.4}{392.4}\times100$
$\boxed{\text{percentage deviation}=-6.78\%}$
