Answer

Verified

440.7k+ views

**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.

\[r \propto {n^2}\]

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.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Trending doubts

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Which are the Top 10 Largest Countries of the World?

Difference Between Plant Cell and Animal Cell

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Give 10 examples for herbs , shrubs , climbers , creepers

Change the following sentences into negative and interrogative class 10 english CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE