Question

A mercury atom is initially in its lowest possible (or ground state) energy level. The atom absorbs a photon with a wavelength of 185 nm and then emits a photon with a frequency of 4.924 x 10^14 Hz. At the end of this series of transitions, the atom will still be in an energy level above the ground state. Draw an energy-level diagram for this process and find the energy of this resulting excited state, assuming that we assign a value of E = 0 to the ground state.

Answer 1

Energy of photon absorbed by atom = hc/l

with,

h = planck's constant

c = speed of light

l = 185 nm = 185 x 10^-9 m

we get,

Energy[1] = 6.626 x 10^-34 x 3 x 10^8/185 x 10^-9 = 1.074 x 10^-18 J

Now,

energy of photon emitted = hv

with,

v = 4.924 x 10^14 Hz

we get,

Energy[2] = 6.626 x 10^-34 x 4.924 x 10^14 = 3.26 x 10^-19 J

energy of excited state = Energy[1] - Energy[2] = 7.48 x 10^-19 J

a representative diagram is shown below,