Question

(a) Using the Bohr’s model calculate the speed of the electron in ahydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbitalperiod in each of these levels.

Answer 1

(a)

The orbital speed of a hydrogen atom is given as,

v 1 = e 2 n×4π ε 0 ( h 2π ) (1)

where, permittivity of free space is ε 0 , charge on an electron is e, plank’s constant is h, and mass of electron is m.

For level n=1,

Substitute the values in the above equation.

v 1 = ( 1.6× 10 −19 ) 2 2×8.85× 10 −12 ×6.62× 10 −34 =0.0218× 10 8 =2.18× 10 6  m/s

For level n=2,

v 2 = ( 1.6× 10 −19 ) 2 2×2×8.85× 10 −12 ×6.62× 10 −34 =1.09× 10 6  m/s

For level n=3,

v 3 = ( 1.6× 10 −19 ) 2 3×2×8.85× 10 −12 ×6.62× 10 −34 =7.27× 10 5  m/s

Thus, the orbital speed of electron in hydrogen atom in first, second and third level is, 2.18× 10 6  m/s, 1.09× 10 6  m/sand 7.27× 10 5  m/s.

(b)

The orbital period of electron at level n is given as,

T n = 2π v n × n 2 h 2 ε 0 πm e 2

For level n=1,

T 1 = 2π 2.18× 10 6 × ( 6.62× 10 −34 ) 2 ×8.85× 10 −12 π×9.1× 10 −31 × ( 1.6× 10 −19 ) 2 =1.527× 10 −16  s

For level n=2,

T 2 = 2π v 2 × ( 2 ) 2 h 2 ε 0 πm e 2 = 2π 1.09× 10 6 × ( 6.62× 10 −34 ) 2 ×8.85× 10 −12 π×9.1× 10 −31 × ( 1.6× 10 −19 ) 2 =1.22× 10 −15  s

For level n=3,

T 3 = 2π v 3 × ( 3 ) 2 h 2 ε 0 πm e 2 = 2π 7.27× 10 5 × ( 6.62× 10 −34 ) 2 ×8.85× 10 −12 π×9.1× 10 −31 × ( 1.6× 10 −19 ) 2 =4.11× 10 −15

Thus, the orbital speed of each these levels is 1.527× 10 −16  s, 1.22× 10 −15  sand 4.11× 10 −15  srespectively.