Question

A hydrogen atom rises form its n = 1 state to the n = 4 state by absorbing energy. If the potential energy of the atom in the n = 1 state be -13.6 eV, then calculate

(a) potential energy in the n=4 state,

(b) energy obserbed by the atom in transition,

(c) Wavelength of the emitted radiation if the atom returns to its original state.

Answer 1

Here, \(E_1 = -13.6eV, n = 4, E_4 = ?\)

(a) \(E_4 = -\frac{E_1}{4^2} = \frac{-13.6e V}{16} = -0.85 eV\)

(b) Energy absorbed by the atom in transition form n = 1 to n = 4 is

\(\triangle E = E_4 - E_1 = -0.85 - (-13.6)\)

\(= 12.75 eV\)

(c) \(\triangle E = \frac {hc} \lambda\)

or \(\lambda = \frac {hc}{\triangle E} \)

\(= \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{12.75 \times 1.6 \times 10^{19}}\)

\(= 970 \times 10^{10}m\)

\(= 970\mathring A\)