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#
