I\'m trying to determine the energy level from which an electron fell (n2) to le
ID: 755435 • Letter: I
Question
I'm trying to determine the energy level from which an electron fell (n2) to level 2 (n1) in a Hydrogen atom. The wavelength of the light emitted is 412.0 nm. Using delta E = hc/lamda, I've determined that the energy corresponding to this wavelenth is 4.82 x 10^-19 J. Now I want to use the equation 1/lamda = R(1/n1^2 - 1/n2^2) such that, 1 / 4.12 X10^-7 m = R(1/2^2 - 1/n2^2) I know that R is a constant and I think its value in this case is 1.097 x 10^-2 ... although I don't entirely understand where this comes from. But if that's correct I can simplify the above equation such that 1/n^2 = 0.0288 I don't remember the arithmetic to get out of the fraction to n^2 = ? Please Help! Thank You!!!!Explanation / Answer
value of Rydberg's constant , R=1.097 * 10^7 m-1 It's an experimentally found constant for Hydrogen atom , for use in Rydberg's formula 1/lamda=R(1/n1^2-1/n2^2) lamda=412*10^(-9) (nm = 10^(-9)m) n1=2 (1/lamda)=R*((1/n1^2)-(1/n2^2)) R=1.097 * 10^7 => n2= 6