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# Calculate velocity of electron in first Bohr orbit of hydrogen atom (given, r =${a_0}$).

Last updated date: 18th May 2024
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Hint: To solve this question we need to use the derived formula of Angular Momentum from de Broglie Equation. According to this equation the angular momentum of the electron relative to the nucleus is quantized.

Complete step by step solution:
In order to solve this question first we need to understand the Bohr model. This model was given by Niels Bohr and Ernest Rutherford. According to this model electrons revolve around the nucleus, consisting of protons and neutrons in circular orbits known as energy shells and can only orbit stably in certain fixed circular orbits at a discrete set of distances from the nucleus.
Let us see the formula of Angular Momentum from de Broglie Equation. From this, we can obtain a formula which is used in the calculation of velocity of electrons.
$mvr = \dfrac{{nh}}{{2\pi }}$, n = number of orbit, where m= mass of electron = $9.1 \times {10^{ - 31}}Kg$, v = velocity of electron that we need to calculate, r = radius of the orbit = $0.529 \times {10^{ - 10}}m$, and h = Planck's constant whose value is$6.626 \times {10^{ - 34}}$ $Kg\quad{m^{ 2 }}/sec$.
Now let up solve this question with the above given equation
Given:

n = 1
$h = 6.626 \times {10^{ - 34}}Kg\quad{m^{ 2 }}/sec$
$m = 9.1 \times {10^{ - 31}}Kg$
$r = 0.529 \times {10^{ - 10}}Kg$

Substituting the values in $mvr = \dfrac{{nh}}{{2\pi }}$
$v= \dfrac{{nh}}{{2mr\pi }}$
$v = \dfrac{{1 \times 6.626 \times {{10}^{ - 34}}}}{{2 \times 3.14 \times 9.1 \times {{10}^{ - 31}} \times 0.529 \times {{10}^{ - 10}}}}m/sec \ \\ v = 2.18 \times {10^6}m/\sec \ \\$
Hence the velocity of the electron in the first Bohr orbit of the hydrogen atom is $2.18 \times {10^6}m/\sec$.

Note: Mass used in the above formula is not the molecular weight of hydrogen, it is the mass of protons of hydrogen atom. And r is the radius of the orbit. So, don't confuse among them.