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Group Problems #34 - Solutions
Monday, November 21
Problem 1 The radial wavefunction
An electron in a hydrogen atom is in a 3d state. (a) What are the principle quantum number and angular momentum quantum num- ber for this electron? For a 3d state, we have n = 3 and ` = 2.
(b) What is the radial wavefunction for this electron? (Hint: L^50 = 1)
From the notes, we have:
R(r) =
na 0
(n โ โ 1)! 2 n[(n +)!] eโr/na^0
2 r na 0
)`
L^2 nโ+1โ 1
2 r na 0
where the last parentheses denote the argument of the associated Laguerre poly- nomials, L^2 nโ+1โ 1. For the 3d state, we have n = 3 and ` = 2, so the polynomial term is L^50 = 1. Evaluating the other constants gives:
R 3 , 2 (r) =
3 a 0
eโr/^3 a^0
2 r 3 a 0
L^50 (2)
(3a 0 )^3 /^2
9 a^20
r^2 eโr/^3 a^0 (4) (5)
=
(3a 0 )^3 /^2
5 a^20
r^2 eโr/^3 a^0 (6)
(7) = (A 3 , 2 ) r^2 eโr/^3 a^0 , (8)
where A 3 , 2 is a constant.
(c) What is the most probable radius to find the electron? The most probable radius for the electron can be found by taking the derivative of the probability density, r^2 R^2 (r) with respect to r and setting this to zero. In particular, r^2 R^23 , 2 (r) = (A 3 , 2 )^2 r^6 eโ^2 r/^3 a^0 (9) (10) =โ d(r^2 R^2 ) dr
= (A 3 , 2 )^2
[
6 r^5 eโ^2 r/^3 a^0 + r^6
3 a 0
eโ^2 r/^3 a^0
]
= (A 3 , 2 )^2 r^5 eโ^2 r/^3 a^0
[
2 r 3 a 0
]
Eqn. (13) is true at r = 0, r = โ, and when the expression in square brackets is zero. The first two options are not interesting, thus we have:
6 = 2 r 3 a 0
=โ r = 9a 0. (14)
Problem 2 High angular momentum orbitals in hydrogen
Show that for those orbitals with the largest possible angular momentum, the most probable radius for the electron is quantized. We follow a similar procedure as problem 1(c), starting with the general expression in Eqn. (1). The states with maximum possible angular momentum have ` = n โ 1, so the radial wavefunctions become:
Rn(r) = An eโr/na^0 rnโ^1 =โ r^2 R n^2 (r) = A^2 n r^2 n^ eโ^2 r/na^0 , (15)
where the constant An has all the terms not containing r. As above, now we just need to take the derivative of r^2 R^2 and set it to zero:
d(r^2 R^2 ) dr = A^2 n
[
2 n r^2 nโ^1 eโ^2 r/na^0 + r^2 n
na 0
eโ^2 r/na^0
]
= A^2 n 2 r^2 nโ^1 eโ^2 r/na^0
[
n โ r na 0
]
=โ r = n^2 a 0. (19)
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