- #1

Flucky

- 95

- 1

## Homework Statement

Calculate the wavelength of the n = 4 → 3 transition in

^{4}He

^{+}to an accuracy of 4 significant figures. (R∞=109 737 cm

^{-1}.) (Fine structure effects can be neglected.)

## Homework Equations

[itex] \frac{1}{λ} = \frac{m}{m_e} R_∞ (\frac{1}{n_1^2} - \frac{1}{n_2^2})[/itex]

*where λ is wavelength, m is the reduced mass, and [itex]R_∞[/itex] is Rydberg constant (1.1x10*

^{5}cm^{-1})[itex] m = m_e \frac{m_N}{m_e + m_N} [/itex]

*where m is the reduced mass, [itex]m_e[/itex] 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 n

_{1}to 3 and n

_{2}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*

m

_{e}= 9.11x10^{-31}m

_{N }= 2 protons + 2 neutrons = 2x(1.673x10^{-27}kg) + 2x(1.675x10^{-27}kg) = 6.696x10^{-27}kgSo 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:

[itex]\frac{1}{λ}[/itex] = 5347 cm

^{-1}

∴ λ = 1.870x10

^{-4}cm = 1.870x10

^{-6}m = 1870 nm

However 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.