The ground state energy of hydrogen atom is -13.6 eV.
(i) What is the kinetic energy of an electron in the `2^(nd)` excited state ?
(ii) If the electron jumps to the ground state from the `2^(nd)` excited state, calculate the wavelength of the spectral line emitted.

Explanation

a. The second excited state corresponds to `n = 3`.
Now, `E_(3) = - (13.6)/(n^(2)) = (13.6)/(3^(2)) = - (13.6)/(9) = 1.51 e V`
The kinetic energy of an electron in the second excited state`= - (-1.51 - 1.51 e V`
b. Energy emitted , when the electron jumps from the second excited state to the ground state.
`E = - 1.51 - (-13.6) = 12.09 e V = 12.09 xx 1.6 xx 10^(-19)J`
The wavelength of the spectral line emitted.
lambda `= (hc)/(E) = (6.62 xx 10^(-34) xx 3 xx 10^(8))/(12.09 xx 1.6 xx 10^(-19))`
`= 1.027 xx 10^(-7) m = 1.027 Å`.