Question

For a hydrogen atom in the ground state. determine (a) the most probable location at which to find the electron and (b) the most probable radius at which to find the electron, (c) Comment on the relationship between your answers in parts (a) and (b).

Short answer

a) The most probable location to find the electron is at$r=a_0$

b) The most probable radius to find the electron is at$r=a_0$

c) Most probable location and most probable radius both are same in the case of ground state of hydrogen atom

Step-by-step solution

 Given data

Hydrogen atom is in the ground state

 concept

Hydrogen atoms are the simplest type of atoms and are composed of a proton and an electron only.

Formula Used:

The radial function$R_{1,0}\left(r\right)$at a distance r is given by.

$R_{1,0}\left(r\right)=\frac{1}{{\left(a_0\right)}^{3/2}}2e^{-r/a_0}$

 To determine the most probable location of finding electron

The radial probability P(r) is given by.

$P\left(r\right)=r^2{R^2}_{1,0}\left(r\right)$

Substitute$\frac{1}{{\left(a_0\right)}^{3/2}}2e^{-r/a_0}$for$R_{1,0}\left(r\right)$in the above equation.

$$\begin{gathered} P\left(r\right)=r^2\left(\frac{1}{{\left(a_0\right)}^{3/2}}2e^{-r/a_0}\right) \\ =\frac{4}{a_0^3}{r}^2{e}^{-r/a_0} \end{gathered}$$

For the most probable location, take the derivative of the probability function and equate to 0

$$\begin{gathered} \frac{d}{dr}P\left(r\right)=0 \\ \frac{d}{dr}\left(\frac{4}{a_0^3}{r}^2{e}^{-2r/a_0}\right)=0 \\ \frac{4}{a_0^3}{r}^2{e}^{-2r/a_0}r\left(2-r\frac{2}{a_0}\right)=0 \end{gathered}$$

From the above equation, we get-

$$\begin{gathered} \left(2-r\frac{2}{a_0}\right)=0 \\ \frac{2(a_0-r)}{a_0}=0 \\ {a}_0-r=0 \\ r=a_0 \end{gathered}$$

So, at , $r=a_0$P(r) will be maximum

(b)

Most probable radius is that radius at which radial probability is maximum, take the derivative of the radial probability and equate to 0.

$$\begin{gathered} \frac{d}{dr}P\left(r\right)=0 \\ \frac{d}{dr}\left(\frac{4}{a_0^3}{r}^2{e}^{-2r/a_0}\right)=0 \\ \frac{4}{a_0^3}{r}^2{e}^{-2r/a_0}\left(2-r\frac{2}{a_0}\right)=0 \end{gathered}$$

From the above equation, we get-

$$\begin{gathered} \left(2-r\frac{2}{a_0}\right)=0 \\ \frac{2(a_0-r)}{a_0}=0 \\ {a}_0-r=0 \\ r=a_0 \end{gathered}$$

So, $r=a_0$, is the maximum radius.

(c)

For a hydrogen atom in the ground state the most probable location to find the electron is at and the most probable radius to find the electron is at $r=a_0$. Therefore, the most probable location and most probable radius both are same in the case of ground state of hydrogen atom