Asked by physics
there is a figure that looks kind of like this:
E6--------------------- E= -0.378eV
E5--------------------- E= -0.544eV
E4--------------------- E= -0.850eV
E3--------------------- E= -1.51eV
E2--------------------- E= -3.40eV
the figure shown above shows the enrgy levels for a hydrogen atom. What is the energy of the photon emitted when the electron in a hydrogen atom drops from energy level E5 to energy level E2?
E=E2-E5 = -3.4-(-0.544)
= -3.4 to .54u
= -2.856 eV
is this done correctly?
E6--------------------- E= -0.378eV
E5--------------------- E= -0.544eV
E4--------------------- E= -0.850eV
E3--------------------- E= -1.51eV
E2--------------------- E= -3.40eV
the figure shown above shows the enrgy levels for a hydrogen atom. What is the energy of the photon emitted when the electron in a hydrogen atom drops from energy level E5 to energy level E2?
E=E2-E5 = -3.4-(-0.544)
= -3.4 to .54u
= -2.856 eV
is this done correctly?
Answers
Answered by
bobpursley
No. The energy level of a photon cannot be negative. Difference=highlevelenergy-lowerlevelenergy.
Answered by
physics
so, how would i write it out?
= -0.544 - -3.40
????????
= -0.544 - -3.40
????????
Answered by
GanonTEK
yes -0.544 - (-3.40) = -0.544 + 3.40
There are no AI answers yet. The ability to request AI answers is coming soon!