One of the lines in the emission spectrum of the hydrogen atom has a wavelength of 656.7 nm, how would you Identify the value of n initial for the transition giving rise to this emission?

1 Answer
Jun 3, 2016

n=3

Explanation:

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