Question

The ionization energy of hydrogen atom is 13.6 electron volt. The energy required to excite electron in hydrogen atom from the ground state to the first excited status

Answer 1

Answer:

E= 13.6 ×Z2/n2

= 13.6 ×1/4

= 3.4 ev

Explanation:

1st excited state means 2nd shell.

Answer 2

Explanation:

Ionization energy of hydrogen atom=−13.6Z

2

=−13.6eV

E

ionization

=

λ

hc

λ

ionisation

6.626×10

−34

×3×10

8

=13.6×1.6×10

−19

λ

ionisation

=912×10

−8

m

ionization energy for 1 mole H=E

ionization

×N

A

=13.6×1.6×10

−19

×6.023×10

23

=1310kJ/mole

=1313kJ/mol(approx)