9. Free quantum particle. In Quantum Mechanics, 2m dx2 Schrodinger\'s equation f

Question

9. Free quantum particle. In Quantum Mechanics, 2m dx2 Schrodinger's equation for a free particle in one dimension. In this equation is the time-independent Schrödinger's equation for a free particle in one dimension. In this equation us nr) is the wavefunction of the particle, m is its mass. E is its (kinetic) energy, while is the fundamental Planck constant. (a) Find the specific solution of this equation satisfying the conditions y(0)-0 and "(0) = 1. (b) Find the spatial period of the solution from (a), ie, the particle's wavelength , in terms of (In Quantum Mechanics, is called the de Broglie wavelength.) E, m, and c) Express the de Broglie wavelength of the particle from (b) in terms of linear momentum p rather than energy E, given that E =p2/2m.

Explanation / Answer

given shrodingers equation for free particle
-(h'^2/2m)d^(psi)/dx^2 = E*psi
h' = h/2*pi ( reduced planks constant)

a. so lets assume psi = Asin(kx)
then d(psi)/dt = Akcos(kx)
d^2(psi)/dt^2 = -Ak^2sin(kx)

putting this in the equation above
(h'^2/2m)Ak^2sin(kx) = EA*sin(kx)
(h'^2/2m)k^2 = E
k = sqroot(2mE)/h'

also, psi(0) = Asin(k*0) = 0
psi'(0) = Akcos(k*0) = Ak = 1
so, A = 1/k = h'/sqroot(2mE)

so the solution is phi(x) = (h'/sqroot(2mE))sin(x*sqroot(2mE)/h')

b. wavelength = 2*pi/k = h/sqroot(2mE)
c. E = p^2/2m
   so wavelength = h/sqroot(p^2) = h/p

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