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]
The energy of the photon emitted is given by:
DeltaE=-hc/lambda
please note that a negative sign must be introduced to the energy expression since energy is released.
combining the two equations gives:
-hc/lambda=-R_H*(Z)^2[1/n_f^2-1/n_i^2] (equation 1)
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
plugging the numbers in (equation 1)
-(6.626*10^-34 J.s*2.998*10^8 m/s) /(397.2*10^-9 m)=-2.178*10^-18 J*(1)^2[1/2^2-1/n_i^2]
-(6.626*10^-34 cancel(J).cancel(s)*2.998*10^8cancel(m)/cancel(s)) /(397.2*10^-9 cancel(m))=-2.178*10^-18 cancel(J)*(1)^2[1/2^2-1/n_i^2]
solve for n_i,
n_i= 7
