Answer:
= - 0.38 eV
Explanation:
Using Bohr's equation for the energy of an electron in the nth orbital,
E = -13.6 [tex]\frac{Z^{2} }{n^{2} }[/tex]
Where E = energy level in electron volt (eV)
Z = atomic number of atom
n = principal state
Given that n = 6
⇒ E = -13.6 × [tex]\frac{1^{2} }{6^{2} }[/tex]
= - 0.38 eV
Hope this was helpful.
