Grade 9Modern Physics

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

Varad

7 Years agoGrade 9

1 Answer

Sujit Kumar

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

Varad

1 Answer

Sujit Kumar

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