Question

Use appropriate relationships from chapter 8 of the textbook to determine the wavelength of the line in the emission spectrum of He+ produced by an electron transition from ni = 7 to nf = 3.

I got 1005nm before but it is apparently incorrect:

This is the correct wavelength for the transition in a hydrogen atom, but this species is the He+ cation which has two protons (Z=2) in its nucleus, and will have an entirely different emission spectrum compared to that of a single electron atom with only one proton. Use the equation

En=−Z2RHn2

where En is the energy of the electron in the nth level, Z is the atomic number corresponding to the number of protons in the nucleus (or total nuclear charge), RH is the Rydberg constant, and n is the energy level, to find the energy of the transition in a single electron species with more than one proton. Then, the wavelength can be calculated once the energy of the transition is known.

Answer 1

As you correctly said, we are dealing with He+ ion and not hydrogen atom. He+ ion is hydrogen LIKE in that they both have one electron; but the atomic number of helium is 2 and that must be taken into consideration. To do that we simply need to multiply the usual energy level formula by Z². Make sure to use R_H value = 2.18 X 10^-18 J as energy is in Joules. Substitute all the values in the formula:

Delta E = Z²R_H [(1/nf²) - (1/ni²)]

E₃ - E₇ = Z²R_H [(1/3²) - (1/7²)]

E₃ - E₇ = 4 X 2.18 X 10^-18 J X [(1/9) - (1/49)]

E₃ - E₇ = 8.72 X 10^-18 J (0.0907)

E₃ - E₇ = 7.909 X 10^-19 J

We know the relation -

E = h X frequency

It can also be written in terms of the wavelength.

E = (h X c) / wavelength

wavelength = (h X c) / E

h = Planck's constant in Joules.second and c = speed of light in m/s

wavelength = (6.626 X 10^-34 J.s X 3 X 10⁸ m/s) / 7.909 X 10^-19 J

wavelength = 2.513 X 10^-7 m

Convert the wavelength from metres to nanometres.

We know that 1 m = 1 X 10⁹ nm

wavelength = 251.3 nm