Question

What is the ratio of the Rydberg constant for helium to hydrogen atom?

Answer 1

Rydberg constant for a given isotope is found by multiplying the Rydberg constant for an infinitely-massive nucleus (R∞R∞, a.k.a., “theRydberg constant”) by the ratio of the reduced electron mass for that isotope to the normal electron mass: R=μemeR∞R=μemeR∞.

The reduced electron mass is defined as μe=M∗meme+M=me1+meMμe=M∗meme+M=me1+meM, where MM is the mass of the nucleus, so we can write R=R∞1+meMR=R∞1+meM.

I’m assuming we’re looking for the constants for 11H and 44He.

For most nuclei, the source you’re using won’t have the mass of the nucleus alone, so to find MM, you have to take the mass of a neutral atom and subtract the mass of all the electrons (i.e., M=Matom−Z∗meM=Matom−Z∗me). For the Helium-4 nucleus (which, btw, has the special name of “alpha particle”),

MHe=mα=mHe−2me=4.002603MHe=mα=mHe−2me=4.002603u−2(0.0005485799−2(0.0005485799u)=4.001506)=4.001506u.

I could do a similar calculation for the Hydrogen-1 nucleus, but that’s the mass of a single proton, which, if your reference has the electron mass, it’ll likely have the proton mass, too: MH=mp=1.007276MH=mp=1.007276u

Because we’re taking looking for RHe/RHRHe/RH, we don’t actually need the value of R∞R∞:

RHeRH=11+meMHe(11+meMH)RHeRH=11+meMHe(11+meMH)

=1+meMH1+meMHe=1+meMH1+meMHe

=1+0.00054857991.0072761+0.00054857994.001506=1+0.00054857991.0072761+0.00054857994.001506

=1.000407468

Answer 2

Answer:

1/8 is the required answer