Answer to Question #147059 in Atomic and Nuclear Physics for Sridhar

Question

A hydrogen atom goes to excited state absorbing a photon with wavelength [100/99R]A° where R is the Rydberg constant. If this absorption corresponds to a transition line in the Lyman series calculate the energy of the excited state.
Ans: -0.136ev

Answer 1

The energy of the excited state can be found from

$\displaystyle E_2 - E_1 = h \nu = \frac{hc}{\lambda} = hcR \cdot \frac{99}{100} = R_y \frac{99}{100}$

where $R_y = hc R = 13.605 \;eV$

$E_2 -E_1 = 13.605 \cdot 0.99 = 13.46895 \; eV$

The energy of the ground state of hydrogen atom is $E_1 = -13.605 eV$. Therefore,

$E_2 = 13.46895 + E_1 =13.46895 - 13.605 = -0.13605 \; eV$