Q. The ratio of ionization energy of Bohr's hydrogen atom and Bohr's hydrogen like lithium atom is

1828 169 Atoms Report Error

A

1: 1

16%

B

1: 3

28%

C

1: 9

45%

D

None of these

10%

Solution:

Energy of an electron in the ground state of an atom (Bohr's hydrogen like atom) is given as
$E = - 13.6 Z^{2} e V$
$(Z =$ atomic number of the atom
$\Rightarrow E_{ionisation} = 13.6 Z^{2}$
$\Rightarrow \frac{( E ) _{H}}{( E _{i o n} ) _{L i}} = (\frac{Z _{H}}{Z _{L i}})^{2} = (\frac{1}{3})^{2} = \frac{1}{9}$