Atomic StructureHard
Question
The binding energy of an electron in the ground state of hydrogen-like ions in whose spectrum, the third line of the Balmer series is equal to 108.5 nm, is
Options
A.13.6 eV
B.54.4 eV
C.122.4 eV
D.14.4 eV
Solution
$\frac{1240}{108.5} = B.E.\left( \frac{1}{2^{2}} - \frac{1}{5^{2}} \right) \Rightarrow B.E. = 54.4\text{ eV}$
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