Question

The ionization energy of the hydrogen atom is given to be 13.6 ev. A photon falls on a hydrogen atom which is initially in the ground state and excites it to the n = 4 state. Calculate the wavelength of the photon

Answer 1

Answer:

Explanation:

ΔE=13.6[1n21−1n22]=hcλ  

λ=973.5Å

Answer 2

Answer:

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)