Question

1) (60 points) The ground state of the hydrogen atom: In three dimensions, the radial part of the Schrodinger equation appropriate for the ground state of the hydrogen atom is given by: ke2 -ħ2 d2 (rR) = E(rR) 2me dr2 where R(r) is a function of r. Here, since we have no angular momentum in the ground state the angular-momentum quantum number /=0. (a) Show that the function R(r) = Ae-Br satisfies the radial Schrodinger equation, and determine the values of the energy E and the constant B. Hints: Re-express the equation by defining a = -2and 2me mes, and after performing the differentiations of rR(r) collect the terms that include different powers of r. E = 2 , (b) In three dimensions, the probability density as a function of radius is given by P(r)=r2|R(r). Show that the radius where the probability for finding the electron is maximized is equal to the Bohr radius.

Answer 1

Theme CA) - Ke(ra) = f(48) + d (HR) + MR) +hme Ke² de ² (R) F -hme E (RR) kha 62 DERINE = 2 kere ; for Re ws, () R = =ēxR] LET JRIAL SOL is R = Ae- Bee di (A xe- Pe) + Q A e BH - EA X e Bt ☆ o (42Bx) + a e-B* .- @te e-R* [date [86] = f den + Gdf 1 del & a lé Bt - pré Be) + d e Br - Ere e Ble -Be-px - opea ce-PE. x) + 4 P-PX = - Ērenpe d coope - Be #4 - ре- P* В* * e-B% +а е вх. - Еке- В*

(4- Rp 7e+R+ pexe pt - - Ex e-Pet FIEC OR 14-2P) ePR + (pa tējx e P4 =0 } RADIAL" SCHA ÖD INGER E **0 a= 2 * P = , Ez ap? - Me ke JANS Duo G = hê ê Đ E= - ha 2 me 2me f .-ñ mer ken tlf = - me kże 4 I 2+2 a me ħ 4 ANS Naw, Rctele de-B4 PLH) = 41RCA)? Ę pot pere) = x? TALAė de

PA) = 141 ľnpa ; P (0) = 0 , P(x=00) = 0 Pled FOR MOST PROBABLE PLACE OF POETO FINDING ELECTRON 15 MAXIMISED Det en PCS =0 * [ 1412 4* e*281] =0 HEMO → ! [ ta prapt 7-0 RX e XBX X8 (ap) e Rift-o OR 28=2PR [84 - 4*p] =0 *(1-4B) 10 ^ Hoo I 7 Ivo = me kes - Moore WHICH IS THE BOHR' RADIUS