Question

the orbital speed of the electron in the ground state of hydrogen is v, what will be its orbital speed when it is excited to the state having energy -3.4ev

Answer 1

Velocity at nth orbit is v= (2.165 x10^6) x (Z/n)

actually v= (2( pi ^2)Km(e^2)Z )/(hn)

K= electrostatic constant

m= mass of electron

e= charge of electron

h= planks cnst

n= nth orbit

Z= atomic number

Answer 2

First off,

$E_{o}$ = -13.6 ev

1 ev = 1.602 x $10^{-19}$ J

E = $\frac{E_{o} }{n^{2} }$

given that,

E= -3.4

so,

-3.4 = $\frac{-13.6}{n^{2} }$

1 = $\frac{4}{n^{2} }$

$n^{2}$ = 4

n= 2

As we know,

$V_{n}$ ∝ $\frac{1}{n}$

$V_{n=2}$ = $\frac{v}{2}$

therefore,

speed = $\frac{V}{2}$

Hope this helps :)