What Determines the Emission Line of He II at 468.6 nm?

[nsatya]

1. The emission line of He II at 468.6 nm corresponds to what electronic transition?

2. I know that the formula 1 / λ = R(1 / n_b² - 1/ n_a²) may be useful but I'm not sure how the Rydberg's constant would change since it's helium. Also, I'm not sure what to assume as the initial energy state.

3. I first tried to use the same value for R as it is for hydrogen, but it did not seem to give a reasonable answer. Any help would be greatly appreciated. Thanks!

 

[NateD]

The emission line of He II at 468.6 nm corresponds to the transition from a 3s1 state to a 1s2 state. The Rydberg's constant for helium is different than that for hydrogen and is equal to 18.898 cm-1. Therefore, you can use the formula 1 / λ = R(1/nb2 - 1/na2), with R being equal to 18.898 cm-1 and assuming an initial energy state of 3s1.