Nuclear & Particle Physics - Helium

If a singly ionized Helium atom in an n = 4 state emits a photon of wavelength 470 nanometers, which of the following gives the approximate final energy level E_f  of the atom, and the n value, of n_f  this final state?

	Ef  (eV)	nf
A.	− 6.0	3
B.	− 6.0	2
C.	− 14	2
D.	− 14	1
E.	− 52	1

(GR9677 #40)

Solution:

E_photon = E_i  − E_f 

Helium: 2 electrons, 2 protons, 2 neutron
Singly ionized Helium, He⁺:1 electrons, 2 protons, 2 neutron
He⁺ → Hydrogen-like atom

Bohr's Equation for Hydrogen-like atom: E_n =  −13.6 Z²/n² eV

For Helium, Z = 2,
E_i = E_(n =4)  = − 13.6 (2)² /4² = − 13.6 /4 ≈ − 3.4 eV

E_photon  = hν = hc / λ 
with
h = 6.63 × 10⁻³⁴ Joule.second = 4.1 × 10⁻¹⁵ eV.second
c = 3 × 10⁸ m/s
λ  = 470 nm = 470 × 10⁻⁹ m = 4.7 × 10⁻⁷ m

E_photon = (4.1 × 10⁻¹⁵)(3 × 10⁸) / (4.7 × 10⁻⁷) ≈ 3 eV

E_f  = E_i − E_photon  
=  − 3.4 − 3
= − 6.4  eV

To find n_f :

n_f ² = −13.6 Z²/ E_f 
= −13.6 (2)²/ (− 6.4)
= 54.4/6.4 ≈ 9
n_f  = 3

Answer: A