There are only three bound states for a particle of mass m in a one-dimensional potential well of the form shown in the figure. The depth V0 of the potential satisfies

A. $$\frac{{2{\pi ^2}{\hbar ^2}}}{{m{a^2}}} < {V_0} < \frac{{9{\pi ^2}{\hbar ^2}}}{{2m{a^2}}}$$
B. $$\frac{{{\pi ^2}{\hbar ^2}}}{{m{a^2}}} < {V_0} < \frac{{2{\pi ^2}{\hbar ^2}}}{{m{a^2}}}$$
C. $$\frac{{2{\pi ^2}{\hbar ^2}}}{{m{a^2}}} < {V_0} < \frac{{8{\pi ^2}{\hbar ^2}}}{{m{a^2}}}$$
D. $$\frac{{2{\pi ^2}{\hbar ^2}}}{{m{a^2}}} < {V_0} < \frac{{50{\pi ^2}{\hbar ^2}}}{{m{a^2}}}$$
Answer: Option C
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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