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Examveda

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


This Question Belongs to Engineering Physics >> Quantum Mechanics

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Related Questions on Quantum Mechanics

A particle is placed in a one-dimensional box of size L along the X-axis, (0 < x < L). Which of the following is true?

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