Verify that Eq. 39-44, the radial probability density for the ground state of the hydrogen atom, is normalized. That is, verify that the following is true:0P(r)dr=1

Short Answer

Expert verified

It is proved that 0Prdr=1.

Step by step solution

01

Radial probability density:

The expression of radial probability density is given by,

P(r)=4a3r2e-2ra ….. (1)

Here, a=52.292×10-12mis the Bohr radius.

02

Prove that∫0∞P(r)dr=1:

Integrate the radial probability density from zero to infinity.

0Prdr=04a3r2e-2radr=4a30r2e-2radr=4a3-r2a2e-2ra-a2r2e-2ra-a34e-2ra0=4a3-r22e-2ra-ar2e-2ra-a24e-2ra0

Simplify further:

0Prdr=4a20-0-a24e-2ra0=-4a2a24e-2a-e0=1

Hence, it is proved that .0Prdr=1

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