What is the distance of a #2s# electron from the nucleus of H atom?

It is #6# times the Bohr radius, but you have to suffer through a gauntlet of integration by parts.

The #2s# atomic orbital wave function for #"H"# atom is:

#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 #psi^"*"psi# is the probability density, the average distance from the nucleus, or #<< r >>#, the expectation value for the radial position, is found by solving the following integral:

#<< 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 #int_(0)^(2pi) int_(0)^(pi) sinthetad thetad phi = 4pi#, so...

#= 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"")#