Continuation of the previous problem -rla o The expectation value, (r), for a hydrogen atom in the 2pz orbital can be written 00 1 (²>= 24 √" d² +³ (+-)³² (r)= dr r3 3 0 0 where the integrals over and have already been evaluated and included in this expression. (b) A related quantity, the radial probability density is represented by N r²|R (r) |2, where N is a normalization constant and R(r) for the 2pz orbital is R r √2 (+/-) a R 2P 2 (r) = N -r/2a Note that the "most probable" distance corresponds to the location of the maximum value for the function r²|R(r) |². Find this distance for the 2pz orbital, and report your answer as a multiple of ao, that is, for the 1s orbital, the most probable value is ao, so you would report "1".
Continuation of the previous problem -rla o The expectation value, (r), for a hydrogen atom in the 2pz orbital can be written 00 1 (²>= 24 √" d² +³ (+-)³² (r)= dr r3 3 0 0 where the integrals over and have already been evaluated and included in this expression. (b) A related quantity, the radial probability density is represented by N r²|R (r) |2, where N is a normalization constant and R(r) for the 2pz orbital is R r √2 (+/-) a R 2P 2 (r) = N -r/2a Note that the "most probable" distance corresponds to the location of the maximum value for the function r²|R(r) |². Find this distance for the 2pz orbital, and report your answer as a multiple of ao, that is, for the 1s orbital, the most probable value is ao, so you would report "1".
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Transcribed Image Text:Continuation of the previous problem
The expectation value, (r), for a hydrogen atom in the 2pz orbital can be written
where the integrals over 0 and
(b) A related quantity, the radial
and R(r) for the 2pz orbital is
1
r
2 -rla
0
<^>= 24 - √" ²³ ( - ) ². *
<r)=
dr
e
3
a
0
0
have already been evaluated and included in this expression.
probability density is represented by N r²|R(r) |2, where N is a normalization constant
R
2P
• (-/-)
e
2p
0
(r) =N
-r/2a
Note that the "most probable" distance corresponds to the location of the maximum value for the function r²|R (r) |².
Find this distance for the 2pz orbital, and report your answer as a multiple of ao, that is, for the 1s orbital, the most
probable value is ao, so you would report "1".
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