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

Varad , 6 Years ago
Grade 9
anser 1 Answers
Sujit Kumar

Last Activity: 6 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)

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