For a particle of mass m in a one-dimensional harmonic oscillator potential of the form $${V_{\left( x \right)}} = \frac{1}{2}m{\omega ^2}{x^2},$$ the first excited energy eigen state is $$\psi \left( x \right) = x{e^{ - a{x^2}}}.$$ The value of a is
A. $$\frac{{m\omega }}{{4\hbar }}$$
B. $$\frac{{m\omega }}{{3\hbar }}$$
C. $$\frac{{m\omega }}{{2\hbar }}$$
D. $$\frac{{2m\omega }}{{3\hbar }}$$
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. e-ax2 (e-ax1 - e-ax2)
A. 0.75
B. 0.50
C. 0.35
D. 0.25
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