Question

Help??

Using the conversion factor from eV (electron volts) to joules, determine which energy line for an electron dropping from one energy level to another would show a line at 435 nm in an emission spectrum. Use the formulas and the energy level information below to answer the question.
Needed constants:
1.00eV=1.6*10^(-19) J
C= 3.0*10^(8) M/J
H= 6.63*10^(-34) J*S

Energy Level Values:
E6:E= -0.378eV
E5:E= -0.544eV
E4:E= -0.850eV
E3:E= -1.51eV
E2:E= -3.403V

Answer 1

@Michele_Laino @nincompoop Anyone help?

Answer 2

Use
E=hf
f=E/h

Answer 3

E=difference of 2 energy levels
h=plank's constant
f=frequency of emission

Answer 4

Hey @jim_thompson5910 do you have any idea how to do this? I know he gave me the formula above but im still not 100% sure about what numbers im supposed to plug in. Its like he says one thing, but my book says another so im really confused.

Answer 5

Well I think, since each energy is given in electron volts (eV), you must convert each one of them in joules first. Then we can use the formula that shamin posted and combine it with the speed of light formula in order to solve for the wavelength of each. From there, we will know which one has will be on the 435 nm line :)

\(\sf f=\frac{E}{h}\) ,
E=energy in joules
h= planck's constant
f= frequency

\(\sf \lambda = \frac{c}{f}\)
\(\sf \lambda\) is the wavelength

combine the two you'll get \(\sf \lambda = \frac{hc}{E}\)

Answer 6

For example let the electron drop frm energy level 6 to 5

Answer 7

So
E=E6-E5=?

Answer 8

Lamda=?

Answer 9

Is it 435 nm or not?!

Answer 10

If yes then it is ur answer. But if not u hv to try another

Answer 11

U can let the electron is dropping frm level 6 to level 4 and others more

Answer 12

Thanks you guys. :)

Answer 13

Welcome!

Answer 14

But let me know ur result plz!

Answer 15

What was the answer? @glittergurl0101

Answer 16

I dont know this^
Im sorry