Question

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.

Answer 1

(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.

(B)

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.