The Hydrogenic Atom The Bohr radius is defined by convention not as the mean radius of the electron, but as the 'most likely radius' - the location of the maximum value of the radial distribution function. If a distribution is a differentiable function (which implies continuity), the extrema can be found by identifying locations where the derivative is zero. If a function has multiple extrema, the function must be evaluated at each to identify the largest. Here, we will work with the radial part (R2s(r)) of the wavefunction of the 2s orbital: R2s(r)=21(aZ)3/2(12aZr)eZr/2a Where a40e2h2, and becomes the 'Bohr Radius' when me:a00.529A For this problem, we will consider the case of hydrogen (i.e. Z=1 ), and you can neglect the 0.05% difference and just replace a with a0 at the end. (a) Write down an expression for Pr,2s(r), the radial probability density associated with this wavefunction. Hint: Pay close attention in tutorial when Q5(a)ii is discussed. (b) Determine the location of the extrema of Pr,2s(r) Express these three r 's in terms of a0, but don't yet convert rational expressions to decimals. Hint: These extrema are the three roots of drdPr,2s(r). (We exclude r=0,r, where we can see that the radial function goes to zero.) In your algebra, look to collect a common factor of 2a21r(12ar)er/a once you've taken the derivative. (c) Compare the value of the original radial probability density at the three roots to determine the 'most likely radius' for the 2s state of the hydrogen atom. Express your result in the form rmax2s=Ba0, where B is a decimal constant to three significant figures. Hint: You may wish to write a small script, spreadsheet (Excel/Google/etc.) to help evaluate the function at the numerical values of the two irrational roots. (d) To illustrate that the most likely radius is not necessarily at the mean radius, calculate the average/expectation value of r^ for an electron in the 2s state of the Hydrogen atom. Make use of the following integral identity. 0rnerdr=n+1n