The electron in a hydrogen atom jumps from the orbit with n = 7 to the orbit wit

Question

The electron in a hydrogen atom jumps from the orbit with n = 7 to the orbit with n= 2 Will this jump associated with absorption of energy or emission of energy? What is the principal quantum number corresponding to the initial state of the electron? What is the principal quantum number corresponding to the final state of the electron? Use equation [6.5] from the text book and calculate the energies of the initial and final states. E_i and E_f. Write the formula for change in energy associated with the electron jump Delta E Calculate Delta E associated with the electron jump.in Joules What is the energy that IS absorbed or emitted during the demon jump? find the wavelength (innm) commending to the energy found in question (g)? Use 4 sig figs

Explanation / Answer

a) emission of energy

b) principle quantum number n = 7

c) principle quantum number n = 2

D) Ei = -13.6/n^2 = -13.6/7^2   = -0.28 ev

   Ef = -13.6/n^2 = -13.6/2^2 = -3.4 ev

e) DE = Ef - Ei

f) DE = -3.4-(-0.28) = -3.12 ev

g) energy that is emitted = 3.12 ev

h) E = hc/l

5*10^-19 = (6.625*10^-34*3*10^8/x)

x = 3.975*10^-7 m

= 397 nm

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