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/n_{1}^{2})-(1/n_{2}^{2})]
eV

= -13.6 [(1/1^{2})-(1/6^{2})] 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
10^{23} x 13.22 eV

= 7.96 x 10^{24} eV

Now, we know that 1 eV = 1.6 x 10^{-19}J

So, 7.96 x 10^{24} eV = 7.96 x 10^{24} x 1.6 x
10^{-19}J

= 1.2736 x 10^{6}J

**= 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^{-34}Js

= 211.52 x 10^{-19}/ 6.63 x 10^{-34} Hz

**
= 3.19 x 10 ^{16} Hz** is the frequency of light
emitted when electron falls back to grond state.

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