Question

The product of linear momentum and angular momentum of an electron of the hydrogen atom is proportional to $n^x$, where x is:
(A) 0
(B) 1
(C) -2
(D) 2

Answer 1

The linear momentum and angular momentum of an electron are given by : $mv$ and $mvr$ respectively, we have to multiply them and check the power of n from the equation.

Complete step by step answer:
Angular momentum: $mvr\, = \,\dfrac{{nh}}{{2\pi }}$
Linear momentum: $mv$
Therefore, product of these two:
$mvr\, \times mv\, = \,{m^2}{v^2}r$
And $v\, = \,2.18\,\, \times \,{10^6}\,\dfrac{z}{n}$
$r\, = \,0.53\,\dfrac{{{n^2}}}{z}$
Therefore, putting these values in the product we check the powers of n:
Product is proportional to $\dfrac{1}{{{n^2}}} \times {n^2}\, = 1$
Hence, the product is independent of n, i.e. $n^0$.
The correct option is (a).
Additional information:
The above formulas of angular momentum and velocity of electrons in orbits and radius of the orbits is given by the Bohr’s postulates. Bohr’s model of an atom was given by Neil Bohr to overcome the demerits of Rutherford’s model of an atom. Bohr’s atomic theory was established through Planck’s quantum theory. Bohr’s theory is applicable to the hydrogen atom and all the atoms with a single electron. In such atoms, the electron revolves around a circular orbit around the nucleus of the atom. The centripetal force for the circular motion is given by the electrostatic force between the electron and the nucleus. The electron revolves only in particular orbits that have energy levels equal to one of the energy levels given by Bohr. The electron’s angular momentum can only and only be an integral multiple of $\dfrac{h}{{2\pi }}$. While revolving around the nucleus in these orbits the electrons do not radiate energy.

Note: The students should apply the correct formula of the velocity and radius. The students should not waste their time in calculating the values in the product as only the dependency of n is asked in the question