Question

http://prnt.sc/ci5x80
how do you do this?

Answer 1

i have absolutely no idea.. but it doesnt look like feedback to me

Answer 2

http://prnt.sc/ci5yiv

Answer 3

@ganeshie8 and @Kainui like to do this kind of stuff also @astrophysics

Answer 4

infinite?? lol

Answer 5

ugh, I need to know how to do this by tomorrow lol, do i have to basically try solving for Ni^2?

Answer 6

rhydberg formula something related to energy levels of an atom

Answer 7

alright well, let's solve for n2

Answer 8

\[\frac{ 1 }{ \lambda } = R(\frac{ 1 }{ n^{2}_{f} })-(\frac{ 1 }{ n^{2}_{i} })\]

Answer 9

the answer is 2.18x10^-18

Answer 10

no clue why lol

Answer 11

let me show you how to solve for n^2

Answer 12

@Matt6288 it's better to just solve for the variable you're looking for first then plug your answers in.

Answer 13

http://prnt.sc/ci605o

Answer 14

Step #1

\[\frac{ 1 }{ \lambda }+\frac{ 1 }{ n^{2}_{i} } = R(\frac{ 1 }{ n^{2}_{f} })\]

Answer 15

Step #2

\[(\frac{ 1 }{ \lambda }+\frac{ 1 }{ n^{2}_{i} })*\frac{ 1 }{ R } = \frac{ 1 }{ n^{2}_{f} }\]

Answer 16

help me with this problem, as an example http://prnt.sc/ci60ko

Answer 17

lol I was wasn't even done with the first one

Answer 18

I tried it and I think I got n=4? but honestly I have zero confidence in my answer lol

Answer 19

okay in a sec I want to just finish solving the first one

Answer 20

another concept I want to explain

Answer 21

this is helpful thank you

Answer 22

\[\frac{ 1 }{ \infty^{2} } = 0\]

Answer 23

the bigger your denominator is the smaller your fraction value will be

Answer 24

\[\frac{ 1 }{ 2 } > \frac{ 1 }{ 4 }\]

Answer 25

so as you get to say 1/1000 the fraction just gets smaller.

\[\frac{ 1 }{ 2 } > > \frac{ 1 }{ 1000 }\]

so as you approach infinity well technically the value of that is zero

Answer 26

\[(\frac{ 1 }{ \infty^2}-\frac{ 1 }{ 1 })*c = \Delta~E\]

Answer 27

\[(0-1) = -1*(-2.18*10^{-18})~J = +(2.18*10^{-18})~J\]

Answer 28

but it's positive because you're going from a lower to a higher energy level so the change in energy is going to be positive. think about going up hill.

if you go from a higher to a lower energy level you're going to have negative -energy difference. energy is released.

remember things like to be in a more stable energy level, this involves losing energy. that's why your answer is positive.

Answer 29

Oh I see your other question

n = 4 is the lower energy level

and

lambda is 485 nm

Answer 30

\[\frac{ 1 }{ 7.79s ^{-1}}+\frac{ 1 }{ 4^{2}}=1.096m ^{-1} \times \frac{ 1 }{ n _{f} ^{2}}\]

Answer 31

OMG that took forever to type

Answer 32

thats where im stuck right now..

Answer 33

let me see

Answer 34

how do you flip R to 1/R??

Answer 35

you multiplied both sides by the inverse of R?

Answer 36

You should get an integer because energy levels are quantized

Answer 37

i'm trying to figure out how to solve this

Answer 38

ok ill be right back, i gotta go home haha

Answer 39

\[\frac{ 1 }{ \lambda } = R(\frac{ 1 }{ n^{2}_{f} }-\frac{ 1 }{ n^{2}_{i} })\]

\[\frac{ 1 }{ \lambda*R }+\frac{ 1 }{ n^{2}_{i} } = \frac{ 1 }{ n^{2}_{f} }\]

Answer 40

hmm i'm getting n = 2

Answer 41

I g2g

Answer 42

relaxes means lower energy level

Answer 43

thanks a lot :)

Answer 44

Don't forget to close the question! :)