Atomic structure numerical part-2

Atomic structure

source : chem.Libretexts.org

Atomic structure numerical-

Question 1) Calculate the shortest and longest wavelength in H- spectrum of Lyman series. R_H = 109678 cm^-1 ?

Solution )

For Lyman series n1 = 1

For shortest λ of Lyman series , energy difference in two levels showing transition should be maximum, n2 = ∞

1 / λ =  R_H [( 1/ n₁²) – ( 1/ n₂²)]

1 / λ =  109678  [( 1/ 1²) – ( 1/∞ ²)] 

1/λ =  109678

λ =  0.00000912

λmin = 912 x 10^-8  cm = 912 Å  Ans.

For longest λ of Lyman series , energy difference in two levels showing transition should be minimum, n2 = 2

1 / λ =  R_H [( 1/ n₁²) – ( 1/ n₂²)]

1 / λ =  109678  [( 1/ 1²) – ( 1/2²)] 

1/λ =  109678 x 3 / 4

λ =  4 / 109678 x 3 = 0.00001216

λmax= 1216 x 10^-8  cm = 1216 Å  Ans.

Question 2) Calculate the Rydberg constant if He ⁺ ions are known to have the wavelength difference between the first ( of the longest wavelength ) lines of Balmer and Lyman series equal to 133.7 nm ?

Solution )

For Balmer series , n1 = 2 ,n2 = 3, Z for He = 2

1 / λ1 = Z2 R_H [( 1/ n₁²) – ( 1/ n₂²)]

1 / λ1 =  Z2R_H[( 1/ 2 ²) – ( 1/ 3²)]

λ1 =   36/ 5 Z2 R_H

For Lyman series , n1 = 1 ,n2 = 2, Z for He = 2

1 / λ2 = Z2 R_H [( 1/ 1 ²) – ( 1/ 2²)]

λ2 =   4 / 3 Z² R_H

Given , λ1 – λ2 = 133.7   nm= 133.7  x 10 ^-9 m

putting the values of  λ1 and  λ2,

( 36/ 5 Z ² R_{H ) –} (4/ 3 Z ² R_{H) = 133.7 x 10-9}

(1 /Z ² RH) [ (36/5) – (4/3)] = 133.7  x 10 ^-9

Z for He = 2,

(1 /4 RH) [ (88 / 15)] = 133.7  x 10 ^-9

R _H = 0.01969832 x 10 ⁹ = 1.096 x 10 ⁷ m ^-1

R _H = 1.096 x 10 ⁵  cm ^-1  Ans.

Question 3) Calculate the wavelength and energy of the radiation emitted for the electronic transition from infinite to stationary state of hydrogen atom ?

Solution )

R_H =  1.09678 x 10 ⁷ m ^-1 ,  n1 = 1 , n2 = ∞

1 / λ = R_H [( 1/ n₁²) – ( 1/ n₂²)]

1 / λ =  1.09678 x 10⁷ [( 1/ 1²) – ( 1/∞²)] 

1/λ =  1.09678 x 10⁷

λ =  1 / 1.09678 x 10⁷ = 0.00001216

λ = 9.11 x 10^-8  m  Ans.

E = h c / λ

h = 6.625 x 10 ^-34  J sec. , c = 3 x 10 ⁸ m / sec

E = 6.625 x 10 ^-34 x  3 x 10 ⁸ / 9.11 x 10^-8

Energy = 2.17 x 10 ^-18 J

Question 4) Calculate the radius of Bohr’s third orbit in Li ²⁺ ion ?

Solution )

r_n = 0.529 n ² / Z Å

n = 3 , Z = 3

r₃ =  0.529 x 3 x 3/ 3

r₃ = 1.587 Å Ans.

Question 5) Show that the Balmer series occurs between 3645 Å and 6563 Å . RH = 1.0968 x 10 ⁷ m ^-1.

Solution )

For Balmer series , n1 = 2, n2 = 3 , 4 , 5…….∞

For maximum wavelength , n1 = 2 , n2 = 3

1 / λ max = R_H [( 1/ n₁²) – ( 1/ n₂²)]

          =  1.09678 x 10 ⁷ [( 1/ 2²) – ( 1/3²)] 

          =  1.09678 x 10 ⁷ x 5 / 36

λmax  =  36 / 1.09678 x 10 ⁷  x 5= 6.564 x 10 ⁷ m

λ max = 6564 Å  Ans.

For minimum wavelength , n1 = 2 , n2 = ∞

1 / λ min = R_H [( 1/ n₁²) – ( 1/ n₂²)]

          =  1.09678 x 10 ⁷ [( 1/ 2²) – ( 1/∞²)] 

          =   1.09678 x 10 ⁷ / 4

λmin  =  4  / 1.09678 x 10 ⁷  = 3.647 x 10 ⁷ m

λ min = 3647  Å  Ans.