Calculate the wavelength of the n= 4 -> 3 transition in He

[Flucky]

Homework Statement

Calculate the wavelength of the n = 4 → 3 transition in ⁴He⁺ to an accuracy of 4 significant figures. (R∞=109 737 cm^-1.) (Fine structure effects can be neglected.)

Homework Equations

$\frac{1}{λ} = \frac{m}{m_e} R_∞ (\frac{1}{n_1^2} - \frac{1}{n_2^2})$

where λ is wavelength, m is the reduced mass, and $R_∞$ is Rydberg constant (1.1x10⁵ cm^-1)

$m = m_e \frac{m_N}{m_e + m_N}$

where m is the reduced mass, $m_e$ is mass of an electron, and m_N is the mass of the nucleus.

The Attempt at a Solution

In order to come out with a positive wavelength I set n₁ to 3 and n₂ to 4.

Now the problem I have is with the reduced mass, m. To try and make sense of it I did everything in SI units:

m_e = 9.11x10^-31
m_N = 2 protons + 2 neutrons = 2x(1.673x10^-27 kg) + 2x(1.675x10^-27 kg) = 6.696x10^-27 kg

So that gives:

m = m_e x [itex]\frac{6.696x10^-27}{9.11x10^-31 + 6.696x10^-27}[/itex] = 0.9999m_e

Plugging all the numbers in gives:

$\frac{1}{λ}$ = 5347 cm^-1

∴ λ = 1.870x10^-4 cm = 1.870x10^-6 m = 1870 nmHowever the answer is 468.7 nm which means the reduced mass m should equal 3.99 m_e.

It's only a 1 mark question so I am obviously understanding it all wrong somewhere.

Any help would be very much appreciated.

 

[Fightfish]

You are dealing with helium and not hydrogen. Helium has a nuclear charge of 2. You need to take that into account in modifying your Rydberg formula.

 

[Flucky]

You are dealing with helium and not hydrogen. Helium has a nuclear charge of 2. You need to take that into account in modifying your Rydberg formula.

I had no idea the equation would differ. Just looked up 'Rydberg formula' and a Z² appears. Big thanks, I was beginning to pull my hair out.