Prove that the most probable distance from the proton of an electron in the2sstate of hydrogen is 5.236aB.

Short Answer

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The most probable distance from the proton of an electron in the2s state of hydrogen is 5.236aB.

Step by step solution

01

Given information

We need to prove that the most probable distance from the proton of an electron in the 2s state of hydrogen is 5.236aB.

02

Explanation

We need to prove thatr=5.236aBwhere ris most probable distance.

For that we will find radial probability density for 2sstate:

role="math" localid="1650381765979" Prr=4πr2R2sr2Prr=4πA2s2r21-r2aB2e-raB.

We can treat as constant 4πA2s2r2=Cand rearranging the equation on right hand side ,

Prr=Cr2-r3aB+r44a2Be-raB.

Now we will differentiate it with respect to rdPrdr=C2r-3r2aB+r3aB2e-raB-1aBr2-r3aB+r44aB2e-raBdPrdr=C2-4raB+2r2aB-r34aB2re-raB

Rearranging the terms , we get

0=2-4raB+22aB2-14raB30=2-45.236+25.2362-14

r=5.236aB

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