Identifying the 1093nm Line in Hydrogen Spectra

[hemetite]

Homework Statement

A line of wavelength 1093 nm is observed in the hydrogen spectrum. Identify the transition that leads to this line.

Homework Equations

where n refers to the upper state and n' to the lower state.

1/lamda = R (1/n' - 1/n) -----1

Ei - Ef = hc/lamda ------2

The Attempt at a Solution

i am not very sure with my answer... please help along thanks.

basically i see this as a simultaneous equation with two unknown.
i put lambda = 1093nm in equation 1
hence i get,

0.083401374 = 1/n' - 1/n -------3

this is the part i am not so sure. using the equation 2.

Ei - Ef = hc/lamda

(- 13.606ev / n ' ) - (- 13.606ev / n ) = hc/1093nm

13.606 ev (1/n' - 1/n) = 1.81855 x 10^-19 ------4

If i put eq 3, 1/n' = 0.083401374 + 1/n into eq 4

it will become zero.

Where did i go wrong?

 

[gabbagabbahey]

1/lamda = R (1/n' - 1/n) -----1

Ei - Ef = hc/lamda ------2

The Attempt at a Solution

i am not very sure with my answer... please help along thanks.

basically i see this as a simultaneous equation with two unknown

No, the equations

[tex]\frac{1}{\lambda}=R\left(\frac{1}{n'^2}-\frac{1}{n^2}\right)[/itex] <br /> <br /> and <br /> <br /> $$E_i - E_f =13.606\text{eV}\left(\frac{1}{n'^2}-\frac{1}{n^2}\right) \frac{hc}{\lambda}$$<br /> <br /> (You forgot to squares the "n"s! ) both give you exactly the same information (since $\frac{R}{hc}=13.606\text{eV}$ )<br /> <br /> So you only really have one equation and two unknowns. However, you also know that both $n$ and $n'$ are positive <b>integers</b>, so this should provide you with enough additional information to answer the question.[/tex]