The ionization energy of hydrogen like atom is 9 times of hydrogen atom the radiator of 2nd bohr orbit of this atom

Sujit Kumar
111 Points
3 years ago
The ionization energy of an atom is given by
$E=13.6\frac{Z^{2}}{n^{2}}$
Therefore for Hydrogen atom Z=1 (Atomic Number) ; n=1 (Outermost shell is the first shell)
$\rightarrow E=13.6\ e.v$
Therefore for the hydrogen like atom
$E=13.6*9\ e.v$               (* is the multiplication symbol)
Therefore Z (Atomic Number) of the Hydrogen like atom is 3
Radius of Bohr Orbit is given by
$r=0.529\frac{n^{2}}{Z}$

$\rightarrow r=0.529\frac{2^{2}}{3}$
(n=2 [second bohr orbit] Z=3)
$\rightarrow r=0.80533 \AA$
$(\AA \ is\ 10^{-10} meters)$