Question

Angular momentum of an electron revolving in Bohr's orbit is $\dfrac{2h}{\pi }$ the radius of this Bohr's orbit is
a) 0.529 ${{A}^{o}}$
b) 1.058 ${{A}^{o}}$
c) 2.116 ${{A}^{o}}$
d) 8.464 ${{A}^{o}}$

Answer 1

The atom can be described in terms of quantum numbers that give the total orbital angular momentum and total spin angular momentum of a given state for atoms in the first three rows and the first two columns of the periodic table.

Complete answer:
In free atoms, electrons can only be found in discrete energy states. These sharp energy states are associated with electron orbits or shells in an atom, such as a hydrogen atom. One of the consequences of these quantized energy states is that only certain photon energies are allowed when electrons jump down from higher to lower levels, resulting in the hydrogen spectrum.
The Bohr model correctly predicted the energies of hydrogen atoms, but it had significant flaws that were corrected by solving the Schrodinger equation for hydrogen atoms.
The wavelength associated with the electron in the Bohr model is given by the de Broglie relationship.
$\lambda =\dfrac{h}{mv}$
as well as the standing wave condition that circumference = total number of wavelengths The principal quantum number in the hydrogenic case is n.
$2\pi r=n{{\lambda }_{n}}$
These can be combined to obtain an expression for the electron's angular momentum in orbit. (Note that this assumes a circular orbit, which is generally unwarranted.)
$L=mvr=\dfrac{hr}{\lambda }=\dfrac{hr}{\left[ \dfrac{2\lambda r}{n} \right]}=\dfrac{nh}{2\pi }$
Bohr's angular momentum of an electron is given by $mvr$ or $\dfrac{nh}{2\pi }$(where v is the velocity, n is the orbit in which electron is, m is mass of the electron, and r is the radius of the nth orbit). Bohr's atomic model established several postulates for the arrangement of electrons in various orbits around the nucleus.
The angular momentum of electrons orbiting around according to Bohr's atomic model of the nucleus is quantized. He went on to say that electrons only move in orbits where their angular momentum is an integral multiple of$\dfrac{h}{2}$. Louis de Broglie later explained this postulate regarding the quantization of an electron's angular momentum. A moving electron in a circular orbit, he claims, behaves like a particle-wave.
In the above question,
$L=\dfrac{nh}{2\pi }$
$\dfrac{2h}{\pi }=\dfrac{nh}{2\pi }$
$n=4$
$r=0.537{{n}^{2}}$
$r=0.537\times 16$
$r=8.592{{A}^{o}}$
$\approx r=8.464{{A}^{o}}$
Thus, the answer is option D: $8.464{{A}^{o}}$

Note:
Niels Bohr created the Bohr model in 1913. Electrons exist within principal shells in this model. The atom is depicted in the Bohr model as a central nucleus containing protons and neutrons, with electrons in circular orbitals at various distances from the nucleus.