State Bohr's postulates in connection with structure of atom and derive the relation E n = − R c h n 2 where E n is energy in the n th level, R is Rydberg constant, c is speed of light and n is integer.

Let's consider an electron of mass m charge e revolve around the nucleus of positive charge Z e . The centripetal force is provided by the electrostatic force. m v 2 r = 1 4 πε 0 Z e 2 r 2 m v 2 = Z e 2 4 πε 0 r Divide both side by 2 K . E = m v 2 2 = Z e 2 8 πε 0 r Potential energy, P . E = - Z e 2 4 πε 0 r The energy of the electron in the nth orbit is given by, E = K . E + P . E E = - m Z 2 e 4 8 ε 0 2 h 2 n 2 This equation can be written in the terms of Rydberg constant. E = - Z 2 R h c n 2 Where , R = Rydberg contant R = m e 4 8 ε 0 2 c h 3 c = Speed of light

State Bohr's postulates in connection with structure of atom and derive the relation E n = − R c h n 2 where E n is energy in the n th level, R is Rydberg constant, c is speed of light and n is integer.

Let's consider an electron of mass m charge e revolve around the nucleus of positive charge Z e . The centripetal force is provided by the electrostatic force. m v 2 r = 1 4 πε 0 Z e 2 r 2 m v 2 = Z e 2 4 πε 0 r Divide both side by 2 K . E = m v 2 2 = Z e 2 8 πε 0 r Potential energy, P . E = - Z e 2 4 πε 0 r The energy of the electron in the nth orbit is given by, E = K . E + P . E E = - m Z 2 e 4 8 ε 0 2 h 2 n 2 This equation can be written in the terms of Rydberg constant. E = - Z 2 R h c n 2 Where , R = Rydberg contant R = m e 4 8 ε 0 2 c h 3 c = Speed of light

Write unit of Rydberg constant and derive the relation for wave number 1 λ , 1 λ = Z 2 R 1 n 1 2 − 1 n 2 2 and write the symbols stands for.

The energy of the electron in the nth orbit is given by, E = - m Z 2 e 4 8 ε 0 2 h 2 n 2 Suppose in the excited atom, an electron jumps from some higher energy state n 2 to a lower energy level state n 1 . The energy difference between these states is E 2 - E 1 = m Z 2 e 4 8 ε 0 2 h 2 1 n 1 2 - 1 n 2 2 According to Bohr's third postulate, the frequency of the electromagnetic wave or photon is v = E 2 - E 1 h = m Z 2 e 4 8 ε 0 2 h 3 1 n 1 2 - 1 n 2 2 The corresponding wavelength of the emitted radiation is given by, 1 λ = v c = m Z 2 e 4 8 ε 0 2 c h 3 1 n 1 2 - 1 n 2 2 1 λ = v c = R Z 2 1 n 1 2 - 1 n 2 2 Where , R = m e 4 8 ε 0 2 c h 3 called Rydberg constant The unit of the Rydberg constant is m - 1 . Rydberg constant, R = m e 4 8 ε 0 2 c h 3 Where, e = Electronic charge m = Mass of the electron c = 3 × 10 8 m / s The speed of the light h = 6 . 67 × 10 - 34 Js Planck's constant

The energy of an α -particle is 1 . 2 × 10 - 13 J . Up to what the closest distance can it reach the nucleus of silver Z = 47 ?

Given energy of alpha particle= 1 . 2 × 10 - 13 J Charge number of the target nuclei Z = 47 . Charge on the electron e = 1 . 6 × 10 - 19 C , Value of 1 4 πε 0 = 9 × 10 9 Nm 2 C - 2 The distance of the closest approach of an alpha particle with energy 1 2 m v 2 to a nucleus is given by r 0 = Ze ( 2 e ) 4 πε 0 1 2 mv 2 where Z = atomic number of the nucleus, e = charge on electron. Substituting the values, we have, r 0 = Ze ( 2 e ) 4 πε 0 1 2 mv 2 = 9 × 10 9 × 47 × 2 × 1 . 6 × 10 - 19 2 1 . 2 × 10 - 13 = 1 . 8 × 10 - 13 m

The radius of the first orbit of the electron of a hydrogen atom is 5 . 3 × 10 - 11 m . What is the radius of its second orbit?

The radius of n th electron orbit of hydrogen atom is given by r n = 5 . 29 × 10 - 11 m n 2 Z where n = orbit number and Z = atomic number, for hydrogen this is 1 . It is clear that the radius of orbit is proportional to the square of the number. Therefore, if the radius of first orbit is 5 . 3 × 10 - 11 m then the radius of the second orbit is 5 . 3 × 10 - 11 m × 4 = 21 . 2 × 10 - 11 m

Show that the velocity of electron in the first orbit of hydrogen atom is 1 137 time the velocity of light.

According to the Bohr's atomic model, the centripetal force of the revolving electron is balanced by the electrostatic force between the nucleus ( Ze ) and the electron e , therefore we have, 1 4 πε 0 Ze e r 2 = mv 2 r Where v = velocity, r = radius of orbit and m = mass of the electron, So we have r = 1 4 πε 0 Ze 2 mv 2 ,............................................(1) From quantization of angular momentum, we have, mvr = nh 2 π Or r = nh 2 πmv or v = 2 πZe 2 4 πε 0 nh For hydrogen atom Z = 1 therefore v = e 2 2 ε 0 nh Substituting various values in above equation, we get v = 2 . 2 × 10 6 m / s which is 1 237 of the velocity of light.