Quantum Chemistry/Example 7

Example Problem 7

Using calculus, derive the most probable radius of an electron 1s orbital.

Solution

To determine the equation for the most probable radius of an electron in an 1s orbital we must calculate the value of r at the maximum point on the probability distribution. This is done by setting the derivative of radial probability to zero and solving for radius.

The general formula for the radial probability is.

${\displaystyle {\text{Radial Probabilty}}=P(r)=R_{nl}^{*}(r)R_{nl}(r)\cdot r^{2}=(R_{nl}(r))^{2}\cdot r^{2}}$ [1]

After inserting the wavefunction of a 1s orbital we can simplify the radial probability equation as follows.

 Radial Component of Wave Function of 1s Orbital ${\displaystyle R_{10}(r)=2\left({\frac {Z}{a_{o}}}\right)^{\frac {3}{2}}\cdot e^{{\frac {-Z}{a_{o}}}r}}$

[2]

${\displaystyle P_{1s}(r)=\left(2\left({\frac {Z}{a_{o}}}\right)^{\frac {3}{2}}\cdot e^{{\frac {-Z}{a_{o}}}r}\right)^{2}\cdot r^{2}}$

${\displaystyle P_{1s}(r)=4\left({\frac {Z}{a_{o}}}\right)^{3}\cdot e^{{\frac {-2Z}{a_{o}}}r}\cdot r^{2}}$

${\displaystyle P_{1s}(r)={\frac {4Z^{3}}{a_{o}^{3}}}\cdot e^{{\frac {-2Z}{a_{o}}}r}\cdot r^{2}}$

The derivative of the simplified probability function must then be calculated.

${\displaystyle {\frac {\delta }{\delta r}}\left(P_{1s}(r)\right)={\frac {\delta }{\delta r}}\left({\frac {4Z^{3}}{a_{o}^{3}}}\cdot e^{{\frac {-2Z}{a_{o}}}r}\cdot r^{2}\right)}$

The constants can be excluded from the rest of the equation by moving them to the front of the derivative.

${\displaystyle {\frac {\delta }{\delta r}}\left(P_{1s}(r)\right)={\frac {4Z^{3}}{a_{o}^{3}}}{\frac {\delta }{\delta r}}\left(e^{{\frac {-2Z}{a_{o}}}r}\cdot r^{2}\right)}$

We see that the function is composed of the product of two smaller functions of the radius, multiplied by constants. Therefore we can apply the derivative product rule to solve for the equation of the derivative of the probability function.

 Product Rule of Derivatives ${\displaystyle {\frac {\delta }{\delta r}}\left(f(x)\cdot g(x)\right)={\frac {\delta }{\delta r}}(f(x))\cdot g(x)+f(x)\cdot {\frac {\delta }{\delta r}}(g(x))}$

${\displaystyle {\frac {\delta }{\delta r}}\left(P_{1s}(r)\right)={\frac {4Z^{3}}{a_{o}^{3}}}\left({\frac {\delta }{\delta r}}\left(e^{{\frac {-2Z}{a_{o}}}r}\right)\cdot r^{2}+\left(\left(e^{{\frac {-2Z}{a_{o}}}r}\right)\cdot {\frac {\delta }{\delta r}}r^{2}\right)\right)}$

${\displaystyle {\frac {\delta }{\delta r}}\left(P_{1s}(r)\right)={\frac {4Z^{3}}{a_{o}^{3}}}\left(\left({\frac {-2Z}{a_{o}}}\right)\left(e^{{\frac {-2Z}{a_{o}}}r}\right)\cdot r^{2}+\left(\left(e^{{\frac {-2Z}{a_{o}}}r}\right)\cdot 2r\right)\right)}$

Simplifying and moving the constants now will be beneficial to simplifying the final equation

${\displaystyle {\frac {\delta }{\delta r}}\left(P_{1s}(r)\right)={\frac {8Z^{3}}{a_{o}^{3}}}\left(e^{{\frac {-2Z}{a_{o}}}r}\right)\left(\left({\frac {-Z}{a_{o}}}\right)\cdot r^{2}+r\right)}$

${\displaystyle {\frac {\delta }{\delta r}}\left(P_{1s}(r)\right)=0={\frac {8Z^{3}}{a_{o}^{3}}}\left(e^{{\frac {-2Z}{a_{o}}}r}\right)\left(\left({\frac {-Z}{a_{o}}}\right)\cdot r^{2}+r\right)}$

After setting the derivative of radial probability to zero as seen above it can be seen that the only way that the equation can be equal to zero is if the polynomial part of the formula is equal to zero. This is the case because ${\displaystyle e^{x}\neq 0}$  and the first part of the equation is constant which means it cannot be zero. Therefore... ${\displaystyle 0=\left({\frac {-Z}{a_{o}}}\right)\cdot r^{2}+r}$

The equation for the exact value of the most probable radius in a 1s orbital is then calculated using the quadradic equation.

 Quadratic Equation ${\displaystyle x={\frac {-b\pm \left(b^{2}-4ac\right)^{\frac {1}{2}}}{2a}}}$

Therefore let ${\displaystyle x=r}$ , ${\displaystyle a={\frac {-Z}{a_{o}}}}$ , ${\displaystyle b=1}$  and ${\displaystyle c=0}$  ${\displaystyle r={\frac {-b\pm \left(b^{2}-4ac\right)^{\frac {1}{2}}}{2a}}}$

${\displaystyle r={\frac {-(1)\pm \left((1)^{2}-4\left({\frac {-Z}{a_{o}}}\right)(0)\right)^{\frac {1}{2}}}{2\left({\frac {-Z}{a_{o}}}\right)}}}$

${\displaystyle r={\frac {-1\pm \left(1-0\right)^{\frac {1}{2}}}{\left({\frac {-2Z}{a_{o}}}\right)}}}$

${\displaystyle r={\frac {-1\pm 1}{\frac {-2Z}{a_{o}}}}}$

The most probable radius cannot be a negative distance or zero meaning the numerator must be negative so that it is cancelled out by the negative value of the denominator. This means that the plus or minus operator in the numerator must be minus. ${\displaystyle r={\frac {-1-1}{\frac {-2Z}{a_{o}}}}}$

${\displaystyle r={\frac {1}{\frac {Z}{a_{o}}}}}$

${\displaystyle r=r_{\text{most probable}}={\frac {a_{o}}{Z}}}$

In conclusion the most probable radius for an electron in an 1s orbital is ${\displaystyle r_{\text{most probable}}={\frac {a_{o}}{Z}}}$  where that radius is proportional to the bohr radius ${\displaystyle \left(a_{o}={\frac {4\pi \epsilon _{o}\hbar ^{2}}{m_{e}e^{2}}}=5.2917721090\cdot 10^{-11}m\right)}$  and inversely proportional to the nuclear charge of the nucleus${\displaystyle \left(Z={\text{number of protons in nucleus of atom}}\right)}$ . This means that for an atom with one proton like hydrogen [H] the most probable radius is ${\displaystyle 1\cdot a_{o}=5.2917721090\cdot 10^{-11}m}$  while for the helium ion [He+] which also only has one electron in the 1s orbital the most probable radius is ${\displaystyle 0.5\cdot a_{o}=2.645886055\cdot 10^{-11}m}$ .

1. Engel, Thomas, and Philip Reid. “Chapter 20: The Hydrogen Atom.” Physical Chemistry, 3rd ed., Pearson, Upper Saddle River, 2018, p. 478.
2. Engel, Thomas, and Philip Reid. “Chapter 20: The Hydrogen Atom.” Physical Chemistry, 3rd ed., Pearson, Upper Saddle River, 2018, p. 468.