Question

Calculate the velocity of an electron placed in the third orbit if hydrogen atom. also calculate the number of revolutions per second that this electron makes around the nucleus

Answer 1

Answer: $N\quad =\quad 2.43\quad \times \quad 10^{ 14 }$

For hydrogen atom, velocity of nth orbit is given by,  

                      $V_{ n }=\frac { e^{ 2 } }{ 2hn\epsilon _{ 0 } }$

On simplification, the above equation, it gets reduced to,  

                      $V_{ n }=\frac { c }{ 137n }$

Given, n = 3, c = $3 \times 10^8$ m/s  

                      $V_{ n }\quad =\quad \frac { 3\quad \times \quad 10^{ 8 } }{ 137\quad \times \quad 3 } \quad \\=\quad \frac { 3\quad \times \quad 10^{ 8 } }{ 411 }$

                       $V_{ n }\quad =\quad 7.299\quad \times \quad 10^{ 5 }\quad m/sec$

Number of revolution made by the electrons $=\frac { V_{ n }T }{ 2\pi { R }_{ n } }$

Where,

                        $V_{ n }$ is velocity of electrons in nth orbit and ${ R }_{ n }$ is the radius of nth orbit  

Where,

                         $R_{ n }\quad =\quad \left( 0.53\quad \times \quad 10^{ -10 } \right) n^{ 2 }, i.e. R_{ n }\quad =\quad \left( 0.53\quad \times \quad 10^{ -10 } \right) \quad \times \quad 3^{ 2 }$

                         $R_{ n }\quad =\quad \left( 0.53\quad \times \quad 10^{ -10 } \right) \quad \times \quad 9\quad \\=\quad 4.77\quad \times \quad 10^{ -10 }$

T = 1 sec  

Number of revolutions $=\quad \frac { \left( 7.299\quad \times \quad 10^{ 5 } \right) \quad \times \quad \left( 1 \right)  }{ 2\quad \times \quad 3\quad \times \quad 3.14\quad \times \quad 4.77\quad \times \quad { 10 }^{ 10 } }$

                         $N\quad =\quad \frac { 7.299\quad \times \quad 10^{ 5 }\quad \times \quad 10^{ 10 } }{ 2\quad \times \quad 3.14\quad \times \quad 4.77 }$

                         $N\quad =\quad \frac { 7.299\quad \times \quad 10^{ 15 } }{ 29.9556 }$

                         $N\quad =\quad 2.43\quad \times \quad 10^{ 14 }$

Answer 2

Explanation:

This is answer of this question