What is the change in energy if the electron from Part A now drops to the ground state?
the energy of n = 4.00 in Part A. Now find the energy of the ground state. Then subtract the two energies: ΔE=Efinal−Einitial
-3J?
@SmokeyBrown
She answered her own question? WOah. XD
She put what she thought may be the answer, Howie
Its not right
I don't know what I'm doing wrong, @Vocaloid
Hm. As far as I remember the ground state electron energy is -13.6eV which is -2.179e-18J So whatever the energy was at n = 4, just subtract -2.179e-18J?
Again idk about the sign convention here
So 4-( -2.179e-18J)= 4x10^0 J ?
oh, no, 4 is the energy level, not the energy so you would need to calculate the nrg level at n = 4 using the rydberg equation then subtract the ground level nrg
is the rydberg equation : En==rhc/n^2
Just confused as to which equation it is
yeah that should be the one
En==(1.097x10^7)(6.626x10^-34)(2.998x10^8)/4^2 =1.362x10^-19J
good, then repeat the process w/ n = 1 and then subtract the two energies
okay so do I subtract the n=4 energy from the n=1 or vice versa
final minus initial so n = 1 state minus n = 4 state
oh okay 2.043 x 10^-18
J
I'm a little fuzzy on the sign convention but since it's dropping from a higher nrg level to a lower one it should be decreasing in energy I'd use the negative energy change
okay and 4 sig figs right
depends on what you gave for part A
3?
i gave 1.37
yeah then stay consistent with 3 sig figs
that was the answee for part a
okay
its showing up as wrong
did you give the + or - energy change
both
i tried with both
hm I can probably ask someone else for a second opinion because all i'm getting is 2e-18J
oh duh, I forgot to put the e-18
we got it right thanks
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