Question

(c28p47) The Lyman series for a (new!?) one-electron atom is observed in a distant galaxy. The wavelengths of the first four lines and the short-wavelength limitof this Lyman series are given by the energy level diagram in Figure P28.47. Based on this information, calculate (a) the energies of the ground state and firstfour excited
states for this oneelectron atom.

a. What is the energy of the ground state?

b. What is the energy of the first excited state?

c. What is the energy of the second excited state?

d. What is the energy of the third excited state?

e. What is the energy of the fourth excited state?

f. Calculate the longest-wavelength (alpha) lines in the Balmer series for this atom.

g. Calculate the short-wavelength series limit in the Balmer series for this atom.

Answer 1

energy E =hC/λwhere h is Plank's constant =6.625*10^-34 J sC isspeed of the light =3*10^8m/sthe product of h*C =(6.625*10^-34 J s)(3*10^8m/s) (1A⁰/ 10^-10 m)(1 eV/1.6*10^-19 J)1eV =1.6*10^-19 J1 A⁰=10^-10 m
hc = 12400 eV-A⁰_________________________________________________________________________E_∞- E₁= hc/λ where λ =1520 A⁰ and E_∞=0E₁ = -12400/1520 eVE₁= -8.16 eV is the ground state energy_________________________________________________________________________E₂- E₁= 12400/2026 eVE₂= E₁+12400/2026 eVE₂= -2.04 eV is the energy of the first excited state________________________________________________________________________E₃- E₁= 12400/1709 eV
E₃= -0.902 eV is theenergy of thesecond excited state________________________________________________________________________E₄- E₁= 12400/1621 eVE₄= -0.508 eV is theenergy of thethird excited state________________________________________________________________________E₅- E₁= 12400/1583 eVE₅= -0.325 eV is theenergy of the fourth excited state_________________________________________________________________________let the longest-wavelength (alpha) lines in the Balmer series =λthe transcaction from n =3 state to n =2
E₃- E₂= hc/λλ = 12400/(-0.902+2.04) A
=10896.30931 A
let the short-wavelength series limit in the Balmer series for this atom=λ'E_∞- E₂= hc/λ'λ' =12400/2.04=6078.4313A