Question

What is the ratio of ionisation energy of bohr's hydrogen atom and hydrogen like lithium atom?

Answer 1

relation between Ionisation energy , atomic number and orbit number ,
$\bold{I.E\propto\frac{Z^2}{n^2}}$
Here, I.E is Ionisation energy, Z is atomic no and n is state of orbit number.

For hydrogen , H
Z = 1 , n = 1
Hydrogen like Lithium species is Li⁺⁺
so, Z = 3 and n = 1

Now, $\bold{\frac{I.E_{H}}{I.E_{Li^{++}}}}$ = 1/(3)² = 1/9
Hence answer is 1 :9