Illustration 4:

A doubly ionised Lithium atom is hydrogen like with atomic number 3.

(i) Find the wavelength of radiation required to excite the electron in Li⁺⁺ from the first to the third Bohr Orbit. (Ionization energy of the hydrogen atom equals 13.6 eV).

(ii) How many spectral lines are observed in the emission spectrum of the above excited system?

Solution:

          (i) E_n = –13.6 Z² / n² 

                Excitation energy = ΔE = E₃ – E₁ = –13.6 × (3)² [1/3² – 1/1²] = +13.6 × (9) [1–1/9] = 13.6 × (9) (8/9) = 108.8 eV.

                Wavelength λ = hc / ΔE = (6.63 × 10^–34) (3×10⁸) / (13.6 × 8) (1.6 × 10^–19)

                = (6.63 × (3) / (13.6) (8)(1.16)) 10^–7 

                = 114.26 × 10–10 m 

                = 1143 A

(ii) From the excited state (E₃), coming back to ground state, there can be ³C₂ = 3 possible radiations.

Illustration 5:

Radiation emitted in the transition n = 3 to n = 2 orbit in a hydrogen atom falls on a metal to produce photoelectrons, the electrons emitted from the metal surface with the maximum kinetic energy are allowed to pass through a magnetic field of strength 3.125x10^–4 T and it is observed that that the paths haves radius of curvature 10 mm

Find            (a)    the kinetic energy of the fastest photo electrons.

                (b)    the work function of the metal.

                (c)    the wavelength of the radiation.

Solution:

The kinetic energy of the fastest photoelectrons is given by

         ΔE = hv₂₃ – 13.6 (1/2² – 1/3²) eV

        The Kinetic energy of the fastest photoelectrons is given by

                mv² / r = Bev 

        or     mv = p = Ber

         Kinetic Energy = 1/2 mv²  = B²e²r² / 2m

                = (3.125 × 10–4)2 × (1.6 ×)

                = 0.86 eV

        (a)    The kinetic energy of the fastest electron is 0.86 eV

        (b)    The work function, Φ is given by

                Φ = (1.9 – 0.86) eV = 1.04 eV

        (c)    The wavelength of the emitted radiation is,

                λ = hc/E = (6.63 × 10^–34) × (3×10^–8) / 1.3 × 1.6 × 1^–19 m

                  = 6.54 × 10^{–7 m}

                       = 6540Å