Question

The speed of an electron in first Bohr orbit is $\dfrac{\mathrm{c}}{137},$ The speed of electron in third Bohr orbit of hydrogen will be:
(A) $\dfrac{\mathrm{c}}{137}$
(B) $\dfrac{3 \mathrm{c}}{137}$
(C) $\dfrac{\mathrm{c}}{3 \times 137}$
(D) $\dfrac{9 \mathrm{c}}{137}$

Answer 1

Hint: We know that the Bohr model shows the atom as a small, positively charged nucleus surrounded by orbiting electrons. Bohr was the first to discover that electrons travel in separate orbits around the nucleus and that the number of electrons in the outer orbit determines the properties of an element. The Bohr model can be summarized by the following four principles: Electrons occupy only certain orbits around the nucleus. Those orbits are stable and are called "stationary" orbits. Each orbit has an energy associated with it. The Bohr model is a relatively primitive model of the hydrogen atom, compared to the valence shell atom model. As a theory, it can be derived as a first-order approximation of the hydrogen atom using the broader and much more accurate quantum mechanics and thus may be considered to be an obsolete scientific theory.

Complete step by step answer
We know that an electron is a negatively charged subatomic particle electron that can be either free (not attached to any atom), or bound to the nucleus of an atom. In conductors current flows due to the movement of electrons. The electron, denoted is a fundamental particle with negative electric charge that is found arranged in quantum mechanical orbits about neutral atoms. The electron is a lepton, and therefore has lepton number 1. The antiparticle of the electron is called the positron.
Electron, lightest stable subatomic particle known. It carries a negative charge of $1.602176634 \times 10^{-19}$ coulomb, which is considered the basic unit of electric charge. The rest mass of the electron is $9.1093837015 \times 10^{-31} \mathrm{kg},$ which is only $1 / 1,836^{\text {th}}$ mass of a proton.
$\mathrm{mvr}=\dfrac{\mathrm{nh}}{2 \pi}$
$\mathrm{v}=\dfrac{\mathrm{nh}}{\mathrm{mr}}$
Also $\mathrm{r}=0.526 \mathrm{n}^{2}$
This gives $\mathrm{v}=\dfrac{\mathrm{h}}{2 \pi \mathrm{mn} \times .526}$
So, the $\mathrm{v} \propto 1 / \mathrm{n}$
So when $\mathrm{n}=3, \Rightarrow \mathrm{v}=\dfrac{\mathrm{c}}{(137 \times 3)}$

So the correct answer is option C.

Note: We know that electrons are involved in many applications such as electronics, welding, cathode ray tubes, electron microscopes, radiation therapy, lasers, gaseous ionization detectors and particle accelerators. Electrons are also important for the bonding of individual atoms together. Without this bonding force between atoms matter would not be able to interact in the many reactions and forms we see every day. The other is ionic bonding where an atom gives up electrons to another atom. A proton carries a positive charge (+) and an electron carries a negative charge (-), so the atoms of elements are neutral, all the positive charges cancelling out all the negative charges. Atoms differ from one another in the number of protons, neutrons and electrons they contain.