How much energy in kJ is needed to excite 1 mol of H atoms to the stationary state with quantum number n = 6 from their ground state? B) Calculate the frequency in Hz of emitted light when electrons fall back to their grounds state from their state in part A.
Energy required for this transition
E = -13.6 [(1/n12)-(1/n22)] eV
= -13.6 [(1/12)-(1/62)] eV
= -13.6 [1-(1/36)] eV
= -13.6 [35/36)] eV
= -13.22 eV
So, 13.22 eV energy required for transition of electron from ground state to n=6 stationary state for one atom.
for one mole of H atom the energy required = 6.023 x 1023 x 13.22 eV
= 7.96 x 1024 eV
Now, we know that 1 eV = 1.6 x 10-19J
So, 7.96 x 1024 eV = 7.96 x 1024 x 1.6 x 10-19J
= 1.2736 x 106J
= 1273.6 kJ energy is required for the required transition.
Energy per transition, E = 13.22 x 1.6 x 10-19 J
h = 211.52 x 10-19 J, where h = Planck's constant = 6.63 x 10-34Js
= 211.52 x 10-19/ 6.63 x 10-34 Hz
= 3.19 x 1016 Hz is the frequency of light emitted when electron falls back to grond state.
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