Please Help. I have most of the questions done, but my teacher doesn\'t post ans
ID: 3895108 • Letter: P
Question
Please Help. I have most of the questions done, but my teacher doesn't post answers do I am not sure if they're correct. Thank you so much.
The diagram to tire right shows an energy level diagram for the hydrogen atom. Several transitions are shown and labeled by letters. Which of the lettered transitions correspond to a photon being absorbed by the hydrogen atom? Which of these transitions are in the visible region? Which of the lettered transitions correspond to a photon being emitted by the hydrogen atom? Which of these transitions are in the visible region? The diagram to the right shows the energy level diagram of element X. What is the ionization energy of element X? An atom in the ground state absorbs a photon, then emits a photon with a wavelength of 1240 run, as shown in the diagram. What conclusion can you draw about the energy of the photon that was absorbed? Explain your reasoning. The n = 3 state of hydrogen has E3 = -1.51 eV. Why is tins energy negative? Clearly justify your reasoning. What is the physical significance of the specific number 1.51 eV? Justify your reasoning.Explanation / Answer
1.)
a) the lettered transitions which correspond to photon being absorced by the hydrogen atom = C,D,E
we have hc = 1239.84 eV nm
For C , energy transition = 1.51 - 0.85 = 0.66 ev
wavelength = hc/e = 1239.84/0.66 = 1878.5454 nm
For D , energy transition = 13.61 - 1.51 = 12.1 ev
wavelength = hc/e = 1239.84/12.1 = 102.466 nm
For E , energy transition = 3.4-1.51 = 1.89 ev
wavelength = hc/e = 1239.84/1.89 = 656 nm
Out of these wavelength of 656 nm is in visible spectrum range.
The transition E is in visible spectrum range.
b) the lettered transitions which correspond to photon being absorced by the hydrogen atom = A,B
we have hc = 1239.84 eV nm
For A , energy transition = 13.61 - 3.4 = 10.21 ev
wavelength = hc/e = 1239.84/10.21 = 121.433 nm
For B , energy transition = 3.4 - 0.85 = 2.55 ev
wavelength = hc/e = 1239.84/2.55 = 486.211 nm
Out of these wavelength of 486.211 nm is in visible spectrum range.
The transition B is in visible spectrum range.
2.)
a) Ionization energy = transition energy from n=1 to n=2 = 4-2 = 2ev
b)
The energy of photn absorbed = 4-1 = 3ev
Reasoning : After absorbing the first photon , the atom goes from the state n = 1 to n =3
3.) The energy is negative because there is an attractive force between nucleus and the electron and we need to supply energy to make the electron come farther from nucleus.