Solved problems on Transitions theory

Problem # 04
i. Find the first ionization energy of potassium.
ii. Find the wavelength, frequency and energy of third line of Balmer’s series ( RH = 1.09 x 10^7m-¹)
iii. If the wavelength of the first number of Balmer’s series is 6563Å. Calculate Reydberg constant and wavelength of the first member of the Lyman series.
iv. Find energy associated with electrons in the quantum number 2.

Solution;
i. E = I.E

    k = 2:8:8:1

    E = -13.6eV
            n²
    E = -13.6 × 1.6×10^-19J
                    4²

      = -1.36 x 10^-19J

   E = E₂ – E₁
   E = 0 - -1.36 x 10^-19 J
   E = 1.39 x 10^-19J

ii. n₁ = 2
     n₂= 3 + 1
      1 = RH[  1   _   1  ]
      λ            n₁²     n₂²

      1 = 1.097×10^7×[  1  _  1  ]
       λ                             2²   4²

      1 = 1.097×10^7×[  1  _  1  ]
       λ                             4    16

  λ =           16           
         1.097×10^7×3

 .: wavelength λ = 4.86 × 10^-7

iii.  λ = 6563A°.

 From;
1 = RH [  1  _  1  ]
λ            n₁²   n₂²

   1    = RH [ 1   _  1  ]
6563            2²    3²

   1    = RH [ 5  ]
6563           36

RH =     36    
         6563×5

RH = 1.09 x 10-³ A-¹
RH = 1.9 x 10-³ x 10^10m-¹
 .: RH = 1.09 x 10^7m-¹

iv. E =?
     n = 2
     E = 13.6eV
                 n²
     E = 13.6×1.6×10^-19J
                        2²
 .:Energy,  E= -5.44 x 10-19J

Problem # 05
The U.V light has a wave length 2950 A°. Calculate its frequency and energy 1A°= 10-¹º
Given, E₄=-1.36×10^-19J
             E₂= -5.44×10^-19J

Solution;
 λ = 2950 A°
 f = c
      λ
 f = 3.0×10^-8
    2950×10^-10

 .: The frequency, f = 1.02x10-² S-¹

Since;
E= hf
E = 6.3 x 10-³⁴Js x 1.02 x 10-² S-¹

 .: The energy, E = 6.426 x 10-36J

Problem # 06
a) If the electron dropped from E₄ to E₂, Find its frequency and wave length.
The energy released is given in the line spectrum below.

Wavelength

i. Which line has the highest frequency and energy?
ii. Which line has the lowest frequency and energy?
iii. Find the energy and frequency of each line in (i) and (ii) above?

b)

State either or not the transition of electrons would occur if the energy supplied is;

i. Greater than E4 – E2
ii. Equal to E4 – E2
iii. Less than E4 – E2
iv. Greater than E3 – E2 but less than E4 – E2
v. Smaller than E4 – E2

Solution;
a) E₄ = -1.36 x 10-19J
    E₂ = -15.44 x 10-19J
    E = E₂ – E₁
    E = E₄ – E₂
    E = (-1.36 x 10^-19) –(-5.44 x 10^-19)
    .: energy,E = 4.08 x 10-¹J
But:
    E = hf
f = ∆E
       h
f = 4.08 x 10-¹J
      6.63× 10^-³⁴Js

 .:The frequency, f = 6.15 x 10¹⁴ S-¹

Since;
 f  = c
       λ
 λ = c 
       f
    = 3.0×10^-8 ms-¹
       6.15×10¹⁴ s-¹
.: wavelength, λ = 4.88 ×10^-7m.

 i) A line of wave length 2030A°
ii) A line of wave length 8092 A°
iii) f =?
     E =?
Line of 2030A°
Since;
 f = c
      λ
f = 3.0 × 10^ -8
     2030 ×10-¹º
 .: Frequency, f  = 1.48 x 10^-15s-¹

Yet;
E = hf
E = 6.3 x 10-³⁴Js x 1.48 x 10^15s-¹
 .: Energy, E=9.32 X 10^-19 J

b)

i) If E > E₄– E₁ results transition of electron from E₁ to above E₄.
ii) If E = E₄ – E₂ transition takes place from E₂ to E₄ respectively.
iii) While for values of E < E₄ – E₂ results transition of electrons from E₂ towards E₄ but cannot reach instead hang between E₂ and E₄.
iv) If E > E₃ – E₂ but < E₄ – E₂results transition of electron from E₂ and above E₃ but cannot reaching E₄.


                           Back/Next