The energy levels of a particle of mass m in a potential of the form \[\begin{gathered} V\left( x \right) = \infty ,\,\,\,\,\,\,\,\,\,\,\,\,x \leqslant 0 \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}m{\omega ^2}{x^2},\,\,\,x > 0 \hfill \\ \end{gathered} \] are given, in terms of quantum number n = 0, 1, 2, 3, . . ., by
A. $$\left( {n + \frac{1}{2}} \right)\hbar \omega $$
B. $$\left( {2n + \frac{1}{2}} \right)\hbar \omega $$
C. $$\left( {2n + \frac{3}{2}} \right)\hbar \omega $$
D. $$\left( {n + \frac{3}{2}} \right)\hbar \omega $$
Answer: Option A
A. In the ground state, the probability of finding the particle in the interval $$\left( {\frac{L}{4},\,\frac{{3L}}{4}} \right)$$ is half
B. In the first excited state, the probability of finding the particle in the interval $$\left( {\frac{L}{4},\,\frac{{3L}}{4}} \right)$$ is half This also holds for states with n = 4, 6, 8, . . . .
C. For an arbitrary state $$\left| \psi \right\rangle ,$$ the probability of finding the particle in the left half of the well is half
D. In the ground state, the particle has a definite momentum
A. (e-ax1 - e-ax2)
B. a(e-ax1 - e-ax2)
C. e-ax2 (e-ax1 - e-ax2)
D. None of the above

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