What is the distance of a #2s# electron from the nucleus of H atom?
1 Answer
It is
The
#psi_(2s) = 1/(4sqrt(2pi))(1/a_0)^(3//2)(2 - (r)/a_0)e^(-r//2a_0)#
The average value in general is given by
#<< x >> = int_"allspace" xp(x)dx# ,where
#p(x)# would be the probability distribution function.
Since
#<< r >> = int_(0)^(2pi) int_(0)^(pi) int_(0)^(oo) r cdot psi_(2s)^"*"psi_(2s) cdot r^2drsin thetad thetad phi#
#= 1/(32pi)(1/a_0)^(3)int_(0)^(2pi) int_(0)^(pi) int_(0)^(oo) [(2 - (r)/a_0)e^(-r//2a_0)]^2 r^3drsin thetad thetad phi#
#= 1/(32pi)(1/a_0)^(3)int_(0)^(2pi) int_(0)^(pi) int_(0)^(oo) (4 - (4r)/a_0 + (r/a_0)^2)r^3e^(-r//a_0) drsin thetad thetad phi#
#= 1/(32pi)(1/a_0)^(3)int_(0)^(2pi) int_(0)^(pi) int_(0)^(oo) (4e^(-r//a_0) - (4r)/a_0e^(-r//a_0) + (r/a_0)^2e^(-r//a_0)) r^3drsin thetad thetad phi#
The integral
#= 1/(8)(1/a_0)^(3) int_(0)^(oo) (4e^(-r//a_0) - (4r)/a_0e^(-r//a_0) + (r/a_0)^2e^(-r//a_0)) r^3dr#
#= 1/(8)(1/a_0)^(3) int_(0)^(oo) 4r^3e^(-r//a_0) - (4r^4)/a_0e^(-r//a_0) + (r/a_0)^2r^3e^(-r//a_0)dr#
#= 1/(8)(1/a_0)^(3) int_(0)^(oo) 4r^3e^(-r//a_0)dr - 1/(8)(1/a_0)^(3) int_(0)^(oo)(4r^4)/a_0e^(-r//a_0)dr + 1/(8)(1/a_0)^(3) int_(0)^(oo)(r/a_0)^2r^3e^(-r//a_0)dr#
I'm not going to go past this though, because it is just ugly integration by parts. Using Wolfram Alpha...
- The first integral gives
#3a_0# . - The second gives
#12a_0# . - The third gives
#15a_0# .
So, the result is:
#color(blue)(<< r >>) = 3a_0 - 12a_0 + 15a_0 = color(blue)(6a_0)#
And to check, someone on stackexchange got a general formula:
#<< r >> = a_0/(2Z)(3n^2 - l(l+1))#
#= a_0/(2 cdot 1) (3(2)^2 - 0(0+1)) = 6a_0# #color(blue)(sqrt"")#