Question

The circumference of the first Bohr orbit in H atom is $3.322 \times {10^{ - 10}}$ m. What is the velocity of the electron of this orbit?

Answer 1

Hint:We have to know that, circle is a distinct round way around the core wherein electrons spin around the core. Three-dimensional space around the core where the likelihood of discovering an electron is greatest is called an orbital. It addresses the movement of an electron in one plane.


Complete step-by-step solution:
We have to know that, the Bohr model recommends that the electrons in molecules are in different energy circles around the core (consider planets circling around the sun). In characterizing these circles of changing energy, Bohr utilized the term energy levels (or shells). The degree of energy that normally involves an electron is called its ground state.
We have to know that the molecule, depicted by the Bohr model as a little, emphatically charged core encompassed by electrons. These electrons move around the core in round circles, comparable in construction to the nearby planetary group, then again, actually electrostatic powers have fascination as opposed to gravity.
When the Bohr’s model formula is given below,
$mvr = \dfrac{{nh}}{{2\pi }}$
(or)
$v = \dfrac{{nh}}{{m2\pi r}}$
Where,
$n = 1$
Circumference is, $2\pi r$ .
H atom is $3.322 \times {10^{ - 10}}$ m.
$v = \dfrac{{nh}}{{2\pi mr}} = \dfrac{{1 \times \left( {6.626 \times {{10}^{ - 34}}Js} \right)}}{{\left( {9.1 \times {{10}^{ - 31}}kg} \right) \times \left( {3.322 \times {{10}^{ - 10}}m} \right)}}$
Therefore, the value of velocity is given below,
$v = 2.19 \times {10^6}m{s^{ - 1}}$ .

Note: As we know that the Bohr model is a generally crude model of the hydrogen molecule, contrasted with the valence shell iota model. As a hypothesis, it tends to be inferred as a first-request estimation of the hydrogen particle utilizing the more extensive and significantly more exact quantum mechanics and in this manner might be viewed as an out of date logical hypothesis.