In the hydrogen atom, the energy of the electron in a given energy level is given by : E_n= -R_H*(Z/n)^2
E_f= -R_H*(Z/n_f)^2
E_i= -R_H*(Z/n_i)^2
DeltaE=E_f-E_i
DeltaE=[-R_H*(Z/n_f)^2]-[ -R_H*(Z/n_i)^2]
take -R_H*(Z)^2 as a common factor,
DeltaE=-R_H*(Z)^2[1/n_f^2-1/n_i^2] " (Eq. 1)"
The energy of the photon emitted is given by:
DeltaE=-hc/lambda " (Eq. 2)"
please note that a negative sign must be introduced to the energy expression since energy is released.
combining the two equations (Eq. 1 and Eq.2) gives:
-hc/lambda=-R_H*(Z)^2[1/n_f^2-1/n_i^2] " (Eq. 3)"
h" is Planck's constant" = 6.626*10^-34 J.s
R_H" is Rydberg constant" = 2.178*10^-18 J
Z" is the atomic number of the hydrogen atom" = 1
n" is principle quantum number"
n_i =?" " is the initial quantum state of the electron.
n_f =2 " " since the wavelength emitted lies in the visible region of the spectrum ( Balmer series). All visible transitions must end up with n= 2.
plugging the numbers in "(Eq.3)"
-(6.626*10^-34 J.sxx2.998*10^8 m/s) /(656.7xx10^-9 m)=-2.178*10^-18 J*(1)^2[1/2^2-1/n_i^2]
-(6.626*10^-34 cancel(J).cancel(s)xx2.998*10^8cancel(m)/cancel(s)) /(656.7xx10^-9 cancel(m))=-2.178*10^-18 cancel(J)*(1)^2[1/2^2-1/n_i^2]
solve for n_i,
n_i= 3