What Is the Rydberg Constant Value for Calculating Hydrogen Atom Transitions?

[craig.16]

Homework Statement

The ionisation energy of a single hydrogen atom is 13.6 eV. The application of Bohr quantisation to the hydrogen atom results in the stationary states having discrete energies given in terms of a positive integer n according to

E=-R_B/n²

where R_B is the Rydberg constant. Determine the value of the Rydberg constant. Assuming like Bohr that, when an atom emits or absorbs radiation, it does so in the form of a single quantum, compute the wavelength of the spectral line of the Balmer series (the transition from n=3 to n=2) for hydrogen.

Homework Equations

E=-R_B/n²
$$\stackrel{1}{\lambda}$$=R_B[$$\stackrel{1}{n₁²}$$-$$\stackrel{1}{n₂²}$$]

The Attempt at a Solution

Ok my main issue with this whole question is which value is the Rydberg constant. Is it:
2.18*10^-18J or 1.097373157*10⁷m^-1 (or both)

If I go with the first value, I can easily obtain that using the first equation as it mentions in the equation "it does so in the form of a single quantum", meaning that n in the equation is equal to 1 and its a simple case of multiplying 13.6 eV by 1.6*10^-19 to convert it to joules giving 2.18*10^-18J.

For the second part however I cannot get a reasonable answer using this value of the Rydberg constant, only with the second one as shown below:

Using first value:
$$\stackrel{1}{\lambda}$$=(2.18*10^-18)[$$\stackrel{1}{4}$$-$$\stackrel{1}{9}$$]
lambda=3.03*10^-19m

Using second value:
$$\stackrel{1}{\lambda}$$=(1.097373157*10⁷)[$$\stackrel{1}{4}$$-$$\stackrel{1}{9}$$]
=656nm (visible)

Im guessing that the constant is both values but I've missed a conversion somewhere. Either that or there is something missing in the second equation to amend this issue with using 2.18*10^-18J.

Finally, how do you do fractions on this properly because as you can see on my post, its separated each fraction from the equation and has somehow enlarged itself. I haven't used this forum that much so I'm still unsure on things like this.

Thanks in advance

 

[vela]

There's the Rydberg constant, which you found correctly to be

$$R_\infty = 1.097\times10^7~\textrm{m}^{-1}$$

and the related Rydberg unit of energy, which is

$$1~\textrm{Ry} = hcR_\infty = 13.6~\textrm{eV}$$

Your first equation uses the energy; your second equation uses the Rydberg constant.

To write the equation

$$\frac{1}{\lambda} = R_B\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$

you'd use the code

\frac{1}{\lambda} = R_B\left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)

sandwiched between the tex tags. You should also be able to right-click on the equation to see the TeX used, though I couldn't get it to work here just now.

 

[craig.16]

Thanks for explaining clearly which equation is for which. I just now went through converting the energy to the rydberg constant (per m) and it turns out it was a lot easier than I initially thought. Also I am very grateful for the code you've displayed regarding fractions on here. It turns out I actually used the wrong symbol in the latex reference, but I've found the real fraction symbol now so it's all good.