# Show that the radius of the orbit in the hydrogen atom varies as , where is the principal quantum number of the atom.

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Hint: Show that the radius of the orbit in the hydrogen atom varies as ${n^2}$, where $n$ is the principal quantum number of the atom.

Complete step by step solution:
We have, according to Bohr’s second postulate,
$mvr = \dfrac{{nh}}{{2\pi }}$ …… (1)
Where,
$m$ indicates mass of the electron.
$v$ indicates velocity of the electron.
$r$ indicates the radius of the orbit in which the electron revolves around the heavy mass called nucleus.
$n$ indicates principal quantum number.
$h$ indicates Planck’s constant.

So, rearranging the equation (1), we have:

$mvr = \dfrac{{nh}}{{2\pi }} \\ v = \dfrac{{nh}}{{2\pi mr}} \\$

In an atom, the centripetal force which is experienced by the electron, is provided by the electrostatic force of attraction present between the electron and the nucleus.

Centripetal force can be expressed by the formula:
$F = \dfrac{{m{v^2}}}{r}$
Where,
$F$ indicates the centripetal force experienced by the electron.
$r$ indicates the distance (radius) between the electron and the nucleus.

Electrostatic force can be expressed by the formula:
$F = \dfrac{{k \times Zq \times q}}{{{r^2}}}$
Where,
$F$ indicates the electrostatic force experienced by the electron.
$Z$ indicates the atomic number of the element.
$q$ indicates the charge of an electron.
$r$ indicates the distance between the electron and the nucleus.

Since, the centripetal force experienced by the electrons is equal to the electrostatic force of attraction present between the electron and the nucleus. So, we can write:

$\dfrac{{m{v^2}}}{r} = \dfrac{{k \times Zq \times q}}{{{r^2}}} \\ \dfrac{{{r^2}}}{r} = \dfrac{{k \times Zq \times q}}{{m{v^2}}} \\ r = \dfrac{{k \times Zq \times q}}{{m{v^2}}} \\$

Substituting, $v = \dfrac{{nh}}{{2\pi mr}}$ in the above equation, we get:

$r = \dfrac{{k \times Zq \times q}}{{m{v^2}}} \\ r = \dfrac{{kZ{q^2}}}{{m{{\left( {\dfrac{{nh}}{{2\pi mr}}} \right)}^2}}} \\ r = \dfrac{{kZ{q^2}}}{{m{{\left( {\dfrac{{nh}}{{2\pi mr}}} \right)}^2}}} \\ r = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}mk{q^2}}} \\$
Since, all the quantities, except $n$, are constants.

Hence, it can be written as:
$r \propto {n^2}$
Radius varies as the square of principal quantum number.

Note: In this problem, you are asked to find the relation of radius and principal quantum number. For this, follow the Bohr’s postulate and find the equation for velocity. Substitute the velocity in the equation of radius to find the relation. It is important to note that all quantities are constant except principal quantum number.