Bohr applied this concept to the line spectra of elements. When elements are exc
ID: 920397 • Letter: B
Question
Bohr applied this concept to the line spectra of elements. When elements are excited they emit radiation at faxed wavelengths He proposed that only certain energy levels are allowed / within the structure of an atom Electrons are I allowed to move between these energy levels The light emitted by the elements is a measure of the energy gap between the two electronic states.
For the hydrogen atom, Delta E=- -R Z 2 [i- ‘1 RH = Rydberg constant 2.178 x 1O 8 J n n Z = nuclear charge = 1 for H, 2 for He 6. Calculate the E for the n = 4 to the n = 2 transition in hydrogen. Where on the EMS would this appear? What does the sign mean? 7. A hydrogen atom in its ground state absorbs light with a wavelength of 102.6 nm. Calculate the energy level of the resulting excited state (n=?)
Explanation / Answer
SOLUTION:
The energy of of the transition is given as
E = RH [1 / [ni2 - 1 / nf2]
RH = Reydberg constant = 2.18 X 10-18
ni = 4, nf = 2
E = 2.18 X 10-18[1 / [42 - 1 / 22] = - 4.087 X 10-18J
Negative sign indicates energy is released.
The energy would appear in VISIBLE part of the EMS. As any transition from n = i to n= 2 belongs to Balmer which lies in visible region.
b) E = hc /
E = 6.62 X 10-34 X 3 X 108 / 102.6 X 10-9 1nm = 10-9m
E = 1.947 X 10-18J
E = 2.18 X 10-18[1 / [12 - 1 / n2]
1.947 X 10-18 = 2.18 X 10-18[1 - 1 / n2]
0.893 = [1 - 1 / n2]
n = 3
Hence electron will be excited to third level