Quantum theory predicts that the Rydberg constant (R infinity) R = me^4/8 epsilo
ID: 938510 • Letter: Q
Question
Quantum theory predicts that the Rydberg constant (R infinity) R = me^4/8 epsilon0^2 ch^3 Where m= mass of an electron, e = charge of an electron, h = Planck's constant, epsilon0 is the permitivity of a vacuum. Obtain these constant values from standard tables and calculate the value of R. Calculate the value of R from theory.. Obtain the value of R in units. Carry the final answer to 7 significant figures. Make sure you show how the units cancel to get the final unit of m^-1. The constants are included on the attached sheet. Make sure that you show your work. The hydrogen emission spectra is purple. Briefly describe why this is purple. why is the line at 656 nm the most intense?Explanation / Answer
1) RH=Rydberg’s constant=me^4/8 eo^2 ch^3=[(9.1095*10^-31 kg) (1.6022*10^-19 C)^4]/[(8*8.8542*10^-12 C2 N-1 m-2)^2 *(2.9979*10^8 ms-1)(6.626*10^-34 Js)^3]
RH=1.0972*10^7 m-1
Unit conversion:
[kg C4/C4 N-2 m-4 ms-1 Js]=[kg/(kg-2 m-2 s4 ) m-4 ms-1 kg m s-2 m s]=m-1
2) The hydrogen emission spectrum there is a series of 4 lines in the visible region. A purple line is at a wavelength of 411 nanometers(nm), which is darker than other lines- dark blue line at a wavelength of 434 nm, green line at 486 nm, and an orange-red line at 656 nm.
3) The red H-alpha spectral line (of wavelength 656 nm) of hydrogen gas of the Balmer series, is the photon released due to the transition of electron from the shell n = 3 to n = 2 of H-atom. As red , is one of the most conspicuous color of all ,so this line is most intense.