For the emission spectrum of Be3+, calculate the lowest wavenumber νˉ (in inverse meters) of light produced by electron transitions between n=2, n=3, and n=4.

I don't want to know the answer, just tell me wat method I need to use to get there... Pls x

8.56*10^6

8.56*10^6

by amar chauhan

Yes but wat concepts did u use???

Hi Anonymous, please be bound by MIT EDX honor code or else you will cause trouble for everyone.

Thank you for your kind corporation.

Jiskha is not part of the MITx platform!!! How the honor code of MITx could be applied to an user of other system?

concepts??

To calculate the lowest wavenumber of light produced by electron transitions for the emission spectrum of Be3+, you would need to use the Rydberg formula. The Rydberg formula is given by:

1/λ = R * (Z^2 / n^2 - Z^2 / m^2)

where λ is the wavelength of the light emitted, R is the Rydberg constant (1.097 × 10^7 m^-1), Z is the atomic number of the element (for Be3+, Z would be 4), and n and m are the principal quantum numbers representing the energy levels of the electron transitions.

In this case, you specifically need to calculate the lowest wavenumber νˉ, which is the reciprocal of the wavelength λ. Therefore, you would need to rearrange the Rydberg formula to solve for νˉ:

νˉ = 1/λ = R * (Z^2 / n^2 - Z^2 / m^2)^-1

Using this formula, you can substitute the appropriate values of Z, n, and m, and perform the calculations to obtain the desired lowest wavenumber νˉ.