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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

ravg=∫P(r)rdr,

has the value 1.5a. In this expression for ravg , 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.

Short Answer

Expert verified

ravg=1.5a

Step by step solution

01

Identification of the given data

The given data is listed below as-

The radial probability density is maximum when r = a

02

The radial probability function

Theradial probability functionfor the ground state is given by-

P(r)=(4r2a3)e-2r/a

Here, r is the radius of Hydrogen atom.

03

To Show that the average value of r, defined as ravg=∫P(r)r dr , has the value 1.5 a   

The average value of is defined as:

ravg=∫0∞rP(r)dr…â¶Ä¦(1)

Now, Pr=4r2a3e-2/a

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

ravg=∫0∞4r3a3e-2r/adr

Now, let x=2ra

dx=2dra

ravg=∫0∞ax34a3e-xadx

Solving the above equation,

ravg=a4∫0∞x3e-xdx

Now, ravgwill become ravg.=a(3!)4

Therefore, it is shown that ravg=1.5a.

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