Question 705051

SCQHARD

The Hamiltonian of a particle of mass mm is given by H=p22m+V(x)H = \frac{p^2}{2m} + V(x), with V(x)={−αxfor x≤0βxfor x>0V(x) = \begin{cases} -\alpha x & \text{for } x \le 0 \\ \beta x & \text{for } x > 0 \end{cases} where α,β\alpha, \beta are positive constants. The nthn^{th} energy eigenvalue EnE_n obtained using WKB approximation is En3/2=32(ℏ22m)1/2π(n−12)f(α,β)E_n^{3/2} = \frac{3}{2} (\frac{\hbar^2}{2m})^{1/2} \pi (n - \frac{1}{2}) f(\alpha, \beta). The function f(α,β)f(\alpha, \beta) is

(A)

αβ2(α2+β2)\frac{\alpha\beta}{2(\alpha^2+\beta^2)}

(B)

αβα+β\frac{\alpha\beta}{\alpha+\beta}

(C)

α+β4\frac{\alpha+\beta}{4}

(D)

12α2+β22\frac{1}{2}\sqrt{\frac{\alpha^2+\beta^2}{2}}

Detailed Solution

Using WKB approximation ∫−E/αE/β2m(E−V(x))dx=(n−12)πℏ\int_{-E/\alpha}^{E/\beta} \sqrt{2m(E - V(x))} dx = (n - \frac{1}{2}) \pi \hbar. Substituting the potential, the integral becomes 2m[∫−E/α0E+αxdx+∫0E/βE−βxdx]\sqrt{2m} [\int_{-E/\alpha}^0 \sqrt{E+\alpha x} dx + \int_0^{E/\beta} \sqrt{E-\beta x} dx]. Evaluating this gives 2m[23αE3/2+23βE3/2]=(n−12)πℏ\sqrt{2m} [\frac{2}{3\alpha} E^{3/2} + \frac{2}{3\beta} E^{3/2}] = (n - \frac{1}{2}) \pi \hbar. Solving for E3/2E^{3/2} yields f(α,β)=αβα+βf(\alpha, \beta) = \frac{\alpha\beta}{\alpha+\beta}.

Free Exam

Boost Your Exam Preparation!

Move beyond just reading solutions. Access our comprehensive Test Series, original Mock Tests, and interactive learning modules. Many premium tests are completely free!

  • Original Mocks & Regular Test Series
  • Real NTA-like Interface with Analytics
  • Many Free Tests Available