Assume a hydrogen atom is in the state 2p with m=0, i.e. n = 2, l= 1, m = 0. The

Question

Assume a hydrogen atom is in the state 2p with m=0, i.e. n = 2, l= 1, m = 0. The following questions require some mathematical derivation ? please show your work: 3a) Show that the most probable distance r from the origin (i.e., from the proton which is the nucleus) to find the electron at is 4 times the Bohr radius, ao. 3b) [Extra Credit] Show that the most probable polar angle theta between the radius vector from the origin pointing to the electron and the z-axis is roughly 35 degrees. Hint: The probability to find the electron within some radial distance r...delta r from the origin and at some polar angle theta... delta theta (regardless of angle phi) is given by

Explanation / Answer

R21 = 1/sqrt(3) (1/2a)^(3/2) ( r/a) e^(- r/2 a)

so

P(r) = 4 pi r^2 R^2 = 4 pi r^2/3 (1/2a)^3 (r/a)^2 e^(-r/a)

P(r) = 4 pi/3 (1/2a)^3 (r^4/a^2) e^(-r/a)

max when dP/dr = 0
dP/dr = 4 pi/3 (1/2a)^3*1/a^2* ( 4 r^3 e^(-r/a) - r^4/a e^(-r/a) ) = 0

( 4 r^3 e^(-r/a) - r^4/a e^(-r/a) ) =0

r^3 e^(-r/a) cancels

4 - r/a = 0

r = 4 a

Thus shown

b) now need spherical harmonic Y 1,0 = cos theta

so to maximize theta

P(theta) = sin(theta) Y1,0^2 = sin(theta) cos(theta)^2

max when dP(theta)/dhteta = 0

dP(theta)/dtheta = cos(theta)^3 - 2*sin(theta)^2 cos(theta) = 0

cos(theta)^2 - 2 sintheta^2 = 0

tan(theta)^2 = 1/2
theta = arctan(1/sqrt(2)) = 35.26 degrees

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