Working with wave number equations, use trial-and-error?

Question

I have a wave number question which I have partially solved:

What transition in $\ce{He+}$ ion shall have the same wavenumber as the first line in Balmer series of $\ce{H}$ atom? (a) $7\to5$ (b) $5\to3$ (c) $6\to4$ (d) $4\to2$

Here's how I solved it:

Note: symbol $Z$ is the atomic number
and $1/\lambda$ is wave number (I have written frequency by mistake)

Solving the problem

At this point my teacher told me to continue using a trial and error method, is there any other way to solve this, perhaps using some other equation.

You have one equation with two unknown; there is no way to solve it without eliminating one of the unknowns (e.g. choosing a value of n1 or n2 and working through by trial and error) or needing more data on the transition. — J. LS, May 20 '15 at 11:49
What data about the transition would be sufficient to solve the equation? — Abhishek Mhatre, May 20 '15 at 11:51
I think 5/36 changed to 5/6 in your final steps. And you cancelled off $Z$ with $Z^2$. And the $Z$ is different in both cases. And in the final eqn, you probably forgot to mark them as n1 and n2. And you will have to visit these errors before going for the trial and error method. And thats it. — Mixcels, May 20 '15 at 14:06

score 7 · Answer 1 · edited Nov 22 '17 at 07:30

[…] my teacher told me to continue using a trial and error method […]

But nobody told you to do it manually once you have understood how to adjust the Rydberg constant for $\ce{He+}$ and use the formula to get the energies for the given transistions, in order to compare them with the $\ce{H_{\alpha}}$ line (656.3 nm):

import scipy.constants

transitions = ((7,5), (5,3), (6,4), (4,2))

def get_he_line_nm(n_upper, n_lower):
    '''calculate the energy for a line in a He+ spectrum and return 
       it in nanometer''' 

    energy = 4 * scipy.constants.Rydberg * (pow(n_lower, -2) - pow(n_upper, -2))

    return 1.0/(scipy.constants.nano * energy) 
    # energy in nm


for upper, lower  in transitions:
    wavelength = get_he_line_nm(upper, lower)
    print('{0} -> {1} : {2:7.2f} nm'.format(upper, lower, wavelength))

will give the output

7 -> 5 : 1162.81 nm
5 -> 3 :  320.37 nm
6 -> 4 :  656.11 nm
4 -> 2 :  121.50 nm

This feels like much more effort than the pen and pencil approach — Martin - マーチン, May 22 '15 at 06:45
I agree :-) Once a student has understood the underlying concept and has actually done some calculations by hand to get a feeling for it, namely where to multiply and where to divide, one should encourage them to write a script to automate the task. That leaves more time to nag them with the next problem set :-D — Klaus-Dieter Warzecha, May 22 '15 at 07:00

score 7 · Accepted Answer · edited Nov 13 '16 at 18:00

Yes, there is.

hydrogen: $$E=\frac{1}{hc}R\left(\frac{1}{(n_1)^2}-\frac{1}{(n_2)^2}\right)$$

helium ion: $$E=\frac{1}{hc}R(2)^2\left(\frac{1}{(x_1)^2}-\frac{1}{(x_2)^2}\right)$$

Since the energy is equal, we get: $$\frac{1}{(n_1)^2}-\frac{1}{(n_2)^2} = 4\left(\frac{1}{(x_1)^2}-\frac{1}{(x_2)^2}\right)$$

with simple rearragement: $$\frac{1}{(n_1)^2}-\frac{1}{(n_2)^2} = \frac{1}{\left(\frac{x_1}{2}\right)^2}-\frac{1}{\left(\frac{x_2}{2}\right)^2}$$

Here, one (obviously) possible answer of the equation is: $$n_1 = \frac{x_1}{2}, n_2 = \frac{x_2}{2}$$

First balmer line is $n_1=3, n_2=2$. One possible answer is $x_1=6, x_2=4$

Ok, I know that with single linear equation and two variables, we can get multiple answers in the case of real numbers (actually, infinite). However, quantum number n can only be positive integer. I don't think there are other answers here.

Working with wave number equations, use trial-and-error?

2 Answers2