Chapter 39: Q55P (page 1217)

The radial probability density for the ground state of the hydrogen atom is a maximum when r = a , where is the Bohr radius. Show that the average value of r, defined as
$r_{a v g} = \int P (r) r d r$ ,
has the value 1.5a. In this expression for $r_{a v g}$ , each value of (P)r is weighted with the value of r at which it occurs. Note that the average value of is greater than the value of r for which (P)r is a maximum.

Expert verified

$r_{a v g} = 1.5 a$

Step by step solution

The given data is listed below as-

The radial probability density is maximum when r = a

Theradial probability functionfor the ground state is given by-

$P (r) = (\frac{4 r^{2}}{a^{3}}) e^{- 2 r / a}$

Here, r is the radius of Hydrogen atom.

The average value of is defined as:

$r_{a v g} = \int_{0}^{\infty} r P (r) d r$ ……(1)

Now, $P (r) = (\frac{4 r^{2}}{a^{3}}) e^{- 2 / a}$

Substitute the value of P(r) in equation (1)

$r_{a v g} = \int_{0}^{\infty} (\frac{4 r^{3}}{a^{3}}) e^{- 2 r / a} d r$

Now, let $x = \frac{2 r}{a}$

$d x = \frac{2 d r}{a}$

$r_{a v g} = \int_{0}^{\infty} (\frac{{(a x)}^{3}}{4 a^{3}}) e^{- x} (a d x)$

Solving the above equation,

$r_{a v g} = \frac{a}{4} \int_{0}^{\infty} x^{3} e^{- x} d x$

Now, $r_{a v g}$ will become $r_{a v g} . = \frac{a (3!)}{4}$

Therefore, it is shown that $r_{a v g} = 1.5 a$ .

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