How Do You Calculate Additional Wavelengths in an Emission Spectrum?

[crimsondarkn]

I solved a, but I couldn't figure out b... I've tried different combinations but nothing turned out correctly...

Homework Statement

The emission spectrum of an unknown substance contains lines with the wavelengths 172 nm, 194 nm , and 258 nm, all resulting from transitions to the ground state.

a) Calculate the energies of the first three excited states.

b) Calculate the wavelengths of three other lines in the substance's emission spectrum.

Homework Equations

E = hc/ wavelength

wavelength 1 = 2.58x10^-7 m
wavelength 2 = 1.94x10^-7 m
wavelength 3 = 1.72x10^-7 m

The Attempt at a Solution

a) First Energy level : E = ((6.63x10^-34)(3.00x10^8)) / (2.58x10^-7) = 7.71 x10^-19 J

Convert to eV: 7.71x10^-19 J / 1.6x10^-19 J/eV = 4.82 eV

Second Energy level : E=((6.63x10^-34)(3.00x10^8)) / (1.94x10^-7) = 1.025x10^-18 J

Concert to eV: 1.025x10^-18 J / 1.6x10^-19 J/eV = 6.41 eV

Third Energy level: E = ((6.63x10^-34)(3.00x10^8)) / (1.72x10^-7) = 1.1564x10^-18 J

Concert to eV: 1.1564x10^-18 J / 1.6x10^-19 = 7.22 eV

b) lamda = hc/Ef-Ei

I know it has something to do with subtracting with the 3 energy levels from above to figure out the three new lamdas but that didn't turn out properly.

the answers are: 518nm,782m, and 152 nm.

If anyone could point me to the right direction, I'd be greatly appreciated.

 

[Doc Al]

You've found the energies of the first three excited states: n = 2, 3, & 4. The wavelengths you were given correspond to transitions to the groundstate: from n=2 to n=1, n=3 to n=1, etc.

But transitions do not always have to be to the ground state; they can be from any higher level to any lower level. Using those 3 excited states, what possible transitions exist?

 

[Astronuc]

If one knows the energies of the first three excited states which are 1->g, 2->g, 3->g, then one can find the energy differences 3-1, 3-2, 2-1 similarly and find the corresponding wavelengths.

See - http://hyperphysics.phy-astr.gsu.edu/hbase/hyde.html

http://hyperphysics.phy-astr.gsu.edu/hbase/hyde.html#c4

 

[daveb]

You have three different energy levels each going to the ground state in the first problem, as correctly solved by you. The second part asks for three wavelengths: the wavelengths of photons emitted when the third excited state goes to the second and first excited states, and when the second excited state goes to the first excite state.

 

[crimsondarkn]

Okay, I've done all the calculations but one doesn't give me the right answer... I'm not sure if its the book's error so if u guys could check please..

Three possible transactions: (7.22-4.82) , (7.22-6.41), (6.41-4.82)

wavelength 1 = hc/Ef-Ei = (6.63x10^-34)(3.00x10^8) / (7.22-4.82)(1.6x10^-19) = 5.17x10^-7 m = 517 nm

wavelength 2 = hc/Ef-Ei = (6.63x10^-34)(3.00x10^8) / (6.41-4.82)(1.6x10^-19) = 7.81x10^-7 m = 781 nm

wavelength 3 = hc/Ef-Ei = (6.63x10^-34)(3.00x10^8) / (7.22-6.41)(1.6x10^-19) = 1.534x10^-6 m = (1534 nm?)

Either the book missed the last digit or my answer is wrong since 152 nm is shown in the book for wavelength 3. and yeah..my other answers are off about 1 but that's pretty close ,since we are dealing with such small numbers after all... Thanks again.

 

[daveb]

The book must be wrong since its answer (152 nm) is the shortest wavelength and would have a higher energy than any of the other emissions (shorter wavelength = higher energy).