General Physics: Answer Key # 23 - Edubirdie

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

2024
Views:

384

Pages:

6
Author:

Jabin F.

Prob. A particle of mass m is moving Under the 1-D poth Vin) = ½mwn? 2 Calculate the ground state Energy of the Particle by using H.U.P. Solh p2 H= + mw² 212 2m E= (200) + (17) 23 ΔΔΡ = 2 + mw² (An)² 4 ΔΡ= 2(02) 24 E = 8m(on)² de (122) 0= and d'E d (124) 2 >0 8m(x)3 3 4'm(on)³ (Δη) = 4m²² (02)²= +ve & Real 2 (An)² = + mw ++mw²20x=0 Δη = Real the Y2 Δη = (mw) 2 2. 2 E = + mw²xm 8mx+ 2 mw E = minimum at this value Prob. A particle of mass m moving in I-D Potential Vin) = Vot Br², where vo & β are Constants suitable units. Find the value of minimum energy of particle by using H.U.P. Emin=0 | Emin = Som V(1) = Vo+B212. 2 H=² + vo + βx2 prop Hr f (AP. Dx) Hr (DP)² + Vo +B(0x)² 2m E= H = flon) EH 11 21(02) 2 + B(A4) d(04) I method. V/m) = =+kx² = mw²² Eo = Emin = tw w= Compare. E=H=²+2 2m Emin = tw + Vo + = 0 m Potential V(n) = 2) Vo+, where vo & Bare Constants. find minimum energy of particle by using H.U.P V() = Vo + 2m w²² W 1= 4mw² = 2W 2 3 Emin = ħw' + Vo al 11(11811) = 11 Σ = Sit =((1A1²-161²)() 151 = = 1412 P If=4+2 131 (1A12-1812) 5 = 5+5 Wi = Asxp (ip-let) 7 2 1 5 = 14P = AR m "B'exp - ipx-let 14 29 = 1421²() = 149() (7'rolch = Aexp

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