Velocity of the electron in the first bohr orbit

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Q7) Calculate the velocity of electron in the first Bohr orbit of hydrogen atom. [Given:- Bohr radius = 0.529 Å, h = 6.626×10-34 Js, mass of the electron = 9.1×10-31 kg]

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Answer : The velocity of electron in the first Bohr orbit of hydrogen atom is, 2.189\times 10^6m/s

Explanation :

Formula used for velocity of electrons in Bohr orbit of hydrogen atom is :

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

where,

v = velocity of electron = ?

n = principle quantum number = 1

h = Planck's constant = 6.626\times 10^{-34}Js

m = mass of electron = 9.1\times 10^{-31}Kg

r = Bohr radius = 0.529A^o=0.529\times 10^{-10}m

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

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

v=\frac{1\times 6.626\times 10^{-34}Js}{3\times 3.14\times 9.1\times 10^{-31}Kg\times 0.529\times 10^{-10}m}

v=2.189\times 10^6m/s

Therefore, the  velocity of electron in the first Bohr orbit of hydrogen atom is, 2.189\times 10^6m/s


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