Question

Calculate angular momentum of an electron in the third Bohr orbit of hydrogen atom.

Answer 1

Hint: According to Bohr postulates, the motion of an electron in a circular orbit around the nucleus is restricted in such a way that its angular momentum is an integral multiple of $h/2\pi $.
Then, $mvr = nh/2\pi $, where $m$ is mass of an electron, $v$ is velocity of the electron, $r$ is radius of orbit and $h$ is Planck's constant.
Complete step by step answer:
According to Bohr postulates, the motion of an electron in a circular orbit around the nucleus is restricted in such a way that its angular momentum is an integral multiple of $h/2\pi $.
Then, angular momentum of an electron is $mvr = nh/2\pi $, where $m$ is mass of an electron, $v$ is velocity of the electron, $r$ is radius of orbit and $h$ is Planck's constant.
In the above equation $n$ is the orbit number in which electrons are present.
As given electron is in third Bohr orbit, then $n = 3.$
Hence the angular momentum of an electron in the third Bohr orbit of a hydrogen atom is given by $mvr = \dfrac{{3h}}{{2\pi }}$.

Note: Bohr’s postulates are only applicable on Hydrogen like atoms means atoms with only one electron in their valence shell. Bohr’s gives the postulates about hydrogen atoms but these are also applicable hydrogen like atoms. According to Bohr an atom has a number of stable orbits in which an electron can reside without the emission of radiant energy and each orbit corresponds to a particular energy level.