JEE MAIN - Physics (2003 - No. 6)
If the binding energy of the electron in a hydrogen atom is $$13.6eV,$$ the energy required to remove the electron from the first excited state of $$L{i^{ + + }}$$ is
$$30.6$$ $$eV$$
$$13.6$$ $$eV$$
$$3.4$$ $$eV$$
$$122.4$$ $$eV$$
설명
$${E_n} = - {{13.6} \over {{n^2}}}{Z^2}eV/$$atom
For lithium ion $$Z=3;$$ $$n=2$$ (for first excited state)
$${E_n} = - {{13.6} \over {{2^2}}} \times {3^2} = - 30.6eV$$
For lithium ion $$Z=3;$$ $$n=2$$ (for first excited state)
$${E_n} = - {{13.6} \over {{2^2}}} \times {3^2} = - 30.6eV$$