Question

The radius of the first Bohr orbit of a hydrogen atom is $0.53\times {10}^{-8}cm.$ The velocity of the electron in the first Bohr orbit is:

• A. $2.188\times {10}^{8}cm{s}^{-1}$
• B. $4.376\times {10}^{8}cm{s}^{-1}$
• C. $1.094\times {10}^{8}cm{s}^{-1}$
• D. $2.188\times {10}^{9}cm{s}^{-1}$

Answer 1

Answer: (A)

$2.188\times {10}^{8}cm{s}^{-1}$

The velocity of electron in the first Bohr orbit of a hydrogen atom is,$2.18\times \frac{{10}^{6}m}{s}$

The formula used for the velocity of electrons in Bohr orbit of a hydrogen atom is :

$v=\frac{nh}{3\pi mr}$

where,

v = velocity of electron = ?

n = principle quantum number = 1

h = Planck's constant =

m = mass of electron =

r = Bohr radius =

Conversion used : $1A^0={10}^{-10}m$

Now put all the given values in the above formula, we get :

$\Rightarrow v=\frac{1\times 6.6\times {10}^{-34}}{2\times 3.14\times 9.1\times {10}^{-31}\times 0.53\times {10}^{-10}}$$=2.18\times {10}^{6}m{s}^{-1}$