Question

Problem 2. Using the Bohr quantization, find approximate values of discrete energy levels for one-dimensional motion of an electron in a symmetric triangular potential well, V(x) αΙΧΙ

Answer 1

According to Bohr quantization, alpha^n middot = -z^2 me^4/ n^2 h^2 epsilon_0 = -13.6 z^2/n^2 therefore a_0 = 0.02529 nm Bohr radius therefore A triangular well consist of potential with a constant slope that is bound at one side by infinite barrier therefore we are given V (x) = alpha Vertical-bar x Vertical-bar therefore alpha middot = C_n {alpha Vertical-bar (x) Vertical-bar hwithstroke Vertical-bar ^2/2m^*}^1/3 C_n = {3/2 pi (n - 1/4)}^2/3 \r\n therefore alpha middot = -1/n^2 m e/2 hwithstroke^2 (e^2/4pi epsilon_0)^2 For psi (x) = squareroot 2/L sin (n pi x/L) For potential becomes, -hwithstroke^2/2m d^2 psi (x)/dx^2 + q epsilon x psi (x) = alpha middot = psi (x) psi (x) For triangular well, psi (x) = A (2m/hwithstroke^2 q^2 epsilon^2)^1/3 {q (epsilon x) -)} where alpha^n middot_n = -(q^2 epsilon^2 hwithstroke^2/2m)^1/3 a_n a_n = -[3pi/2 (n - 1/4)]^2/3 Corresponding energy values are alpha^n middot_n = (hwithstroke^2/2m)^1/3 (3pi q epsilon/2 (n - 1/4))^2/3 with n = 1, 2 ellipsis alpha^n middot