Respuesta :
The minimum wavelength required to remove the electron from the n = 2 level of [tex]Be^{3+}[/tex] is 22.80 nm.
What is ionization energy?
It is the energy needed to remove one electron from a neutral atom, which results in the formation of an ion. The measurement is based on an isolated atom in its gaseous phase and is often expressed in kJ/mol.
d) Now the atom is ionized when the electron is removed from [tex]$n_{\text {initial }}=2$[/tex] to the infinity's shell [tex]$n_{\text {final }}=\infty$[/tex] ). For [tex]$B e^{3+}$[/tex], the Z=4. Using the formula, we get
[tex]$\begin{gathered}\Delta E=2.18 \cdot 10^{-18} J \cdot \frac{1}{4} \cdot 4^{2} \\\Delta E=8.720 \cdot 10^{-18} J\end{gathered}$[/tex]
The wavelength is then calculated as
[tex]$\begin{gathered}\lambda=\frac{h \cdot c}{E}\\=\frac{6.626 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s} \cdot 3 \cdot 10^{8} \mathrm{~m} / \mathrm{s}}{8.720 \cdot 10^{-18} \mathrm{~J}} \\\lambda=2.27959 \cdot 10^{-8} \mathrm{~m}\\=22.7959 \cdot 10^{-9} \mathrm{~m}\\=22.80 \mathrm{~nm}\end{gathered}$[/tex]
So, the minimum wavelength is 22.80 nm.
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