Question

a hydrogen atom initially in ground state absorbs a photon which excites it to n=3 estimate frequency​

Answer 1

Answer:

Explanation:

The initial energy of the electron is E1

E_{1}=-\frac{13.6}{1^{2}}

E1 =-13.6 eV

The energy of the electron when it is excited to level n=4 is E2

E_{1}=-\frac{13.6}{4^{2}}

E2=-0.85 eV

The difference between these two energy levels is equal to the energy of the photon absorbed by the electron.

The energy of the photon \DeltaE = E2 - E1

\DeltaE = -0.85 -(-13.6)

\DeltaE = 12.75 eV

The wavelength of the photon can be calculated using relation

\Delta E=\frac{hc}{\lambda }

hc=1240 eV

\\\lambda =\frac{hc}{\Delta E}\\ \lambda=\frac{1240}{12.57}\\ \lambda=98.6\ nm

\\\nu =\frac{c}{\lambda }\\ \nu =\frac{3\times 10^{8}}{98.6\times 10^{-9}}\\\nu =3.04\times 10^{15}\ Hz

The wavelength and frequency of the photon absorbed by the hydrogen atom are 98.6 nm and 3.04\times1015 Hz respectively.