Answer:
ni, the initial value of n associated with the emission = 6
This is found by the application of the electron energy level, and the Rydberg's equation as can be seen in the following explanation
Explanation:
The known variables are
λ = 93.7 × 10-9m which is a Lyman series as it lies in the UV region
The series is in effect when the final destination of the electron is the first shell, hence
nf = final value of n =1 for ultraviolet light
The energy levels of one electron ions are given by the equation
En = -2.18×10-18J × Z2/n2
Z = atomic number and
n = energy level
Ef = -RH × (Z2/nf2) and Ei = -RH × (Z/ni)2
∆E = Ef - En = -RH × (Z/nf)2 – (-RH × (Z/ni)2)
Collecting like terms we have
∆E = - RH × Z × (1/nf2 – 1/ni2)
For released energy, the emitted photon energy, ∆E, is
∆E = -h×c/λ
h = Planck’s constant = 6.626×10-34J×s
λ = wavelength of the wave = 93.7×10-9m
Z = atomic number of hydrogen = 1
n = principal quantum number
RH = Rydberg constant = 2.178×10-18J
c = speed of light = 299792458 m/s
ni = unknown
Thus we have ∆E = - RH × Z × (1/nf2 – 1/ni2) is the same as
-h×c/λ = - RH × Z × (1/nf2 – 1/ni2)
Inputting the known vales and constants from above we have on the left and right of the above equation
-2.12×10-18J = -2.178×10-18×1×(1/12 – 1/ni2)
Or (1/12 – 1/ni2) = 0.97337, or 1 – 1/ni2 = 0.97337
Which means 1/ni2 = 0.0266 or ni = 6.128 ≈ 6
ni, the initial value of n associated with the emission = 6
