University:
Santa Fe College
Course:
PHY 2048 | General Physics 1 with Calculus
Academic year:

2024
Views:

98

Pages:

5
Author:

Jabin F.

Expectation Value in State In =1/2(3n^2-l(l+1))a_0/z =n^2/2(5n^2+1-3l(l+1)))(a_0/z)^2 <1/r>=1/n^2(a_0) <1/r^2>=2/(n^3(2l+1))z^2/a_0^2 <1/r> ne 1/ ψ(r,o,phi)=1/sqrt(2)(1/a_0)^(3/2)(1-r/(2a_0)e^(-r/2a_0))cos(theta) Here not phi dependence term so m=- here costheta(1 order) l=1 e^(-r)/(2a_0) -> a3 multiply -> n=2 Ques. 3 Page 68 positron → e+ μ = e-e+ E_n = 1 / (4πε_0)^2 * m_e * z^2 / n^2 for positron, z=1 = -13.6 / n^2 eV μ = me+ + me- = me+ + me- = me- + me- = me- P = -me+ / (4πε_0)^2 * m_e * z^2 / n^2 mμ = 1/2 * E_H E_IP = E_H / 2 = -13.6 / 2 = -6.8 eV (option-b) Ques. 2 Ray 72 E_n = 13.6 / n^2 = 13.6 / 2^2 = 6.8 eV (option-b) Same as above Ques. Binding energy = E_H / 2 = 13.6 / 2 = 6.8 eV. Ques. 7. Page-74 [V_{n}=\frac{Z~e^{2}}{4.\Pi\epsilon_{*}n~\hbar}.] Here V is not depends on mass. So for same state of Both atom→ [V_{p=}V_{H}] (option-a) Ques.8. Page-64 for ground state- ψ100 (r) = Ae^-(r/ao) = πa0^3 = ∫ψ100 z ψ100 dτ ∫ψ100 ψ100 dτ = πa0^3 ∫∫∫ e^-(2r/ao) r^3 dr (cosθsinθdθ)dφ = πa0^3 ∫e^-(2r/ao) r^3 dr ∫cosθsinθdθ ∫dφ = 1/πa0^3 ∫e^-(2r/ao) r^3 dr ∫cosθsinθdθ ∫dφ = 0 = = ∫ψn'l'm'* z ψnlm dτ = 0, Otherwise. Prob. Calculate the exp value of of nay (332) Compo. of position of e in ground state of H-atom. Soln ψ100(r) = √(1/(πa0^3)) * e^(-r/a0) x = rsinθcosφ y = rsinθsinφ = (1/(πa0^3)) ∫(0 to ∞) e^(-2r/a0) r^3 dr ∫(0 to π) sin^2θdθ ∫(0 to 2π) cosφ dφ = 0 = (1/(πa0^3)) ∫(0 to ∞) e^(-2r/a0) r^3 dr ∫(0 to π) sin^2θdθ ∫(0 to 2π) sinφ dφ = 0 = (1/(πa0^3)) ∫(0 to ∞) e^(-2r/a0) r^3 dr ∫(0 to π) cosθsinθdθ ∫(0 to 2π) dφ = 0 => = = = 0 => = + + ≠ 0 = √( + + ) = 3a0/2 Prob An e is in the ground state of H-Atom. Where normalizd wave fn is given by ψ100(r) = √(1/(πa0^3)) * e^(-r/a0) Where a0 is first Bohr radius and origin is taken at proton.